7>7<[
/y§i No. 7L//6
POLISH SPACES AND ANALYTIC SETS
THESIS
Presented to the Graduate Council of the
University of North Texas in Partial
Fulfillment of the Requirements
For the Degree of
MASTER OF SCIENCE
By
Kimberly Muller, B.S.
Denton, Texas
August, 1997
Muller, Kimberly, Polish Spaces and Analytic Sets. Master of Science (Mathe-
matics), August, 1997, 81 pp., references, 4 titles.
A Polish space is a separable topological space that can be metrized by means
of a complete metric. A subset A of a Polish space X is analytic if there is a Polish
space Z and a continuous function f : Z —y X such that f(Z^ = A. The idea of
an analytic set was introduced by Souslin and Lusin and was developed mainly by
the Polish and Russian schools of mathematics. We use Polish spaces and analytic
sets to study measure spaces. We show that every Borel subset of a Polish space is
analytic and that every analytic set is universally measurable. After proving that
each uncountable Polish space contains a non-Borel analytic subset we conclude
that there exists a universally measurable non-Borel set.
7>7<[
/y§i No. 7L//6
POLISH SPACES AND ANALYTIC SETS
THESIS
Presented to the Graduate Council of the
University of North Texas in Partial
Fulfillment of the Requirements
For the Degree of
MASTER OF SCIENCE
By
Kimberly Muller, B.S.
Denton, Texas
August, 1997
TABLE OF CONTENTS
Page
Chapter
1. INTRODUCTION !
2. POLISH SPACES 3
2.1 Countablility Axioms 3
2.2 Disjoint Unions n
2.3 Product Spaces
2.4 Each Polish Space is a G$ 25
2.5 Borel Subsets of Polish Spaces and Measurable Spaces 33
3. ANALYTIC SETS 47
3.1 Borel Subsets of Polish Spaces 47
3.2 The Space Af 49
3.3 Zero-Dimensional Spaces 54
3.4 Non-Borel Analytic Sets 5g
4. THE SEPARATION THEOREM AND ITS CONSEQUENCES 66
4.1 Borel Measurable Functions 66
4.2 Borel Isomorphisms 71
5. THE MEASURABILITY OF ANALYTIC SETS 74
REFERENCES 8 1
CHAPTER I
INTRODUCTION
In both analysis and topology, many of the spaces most commonly studied are
separable and metrizable. Often, the metric is also complete. The study of such
spaces, historically called Polish spaces, has led to many interesting results. This
paper is a study of Polish spaces and analytic sets. Many definitions are needed to
begin this study. We say that a sequence (xn)n in a metric space X is a Cauchy
sequence, if for every e > 0 there exists an integer N such that d(xm,xn) < e if
m,n > N. A metric space is said to be complete, if every Cauchy sequence is
convergent. The space of real numbers with the standard metric is an example of a
complete metric space. However, the open interval (0,1) with the standard metric is
not complete because (~)n is a n example of a sequence which is Cauchy but which
does not converge to a point in (0,1). We will find later that it is possible to find
another metric which metrizes (0,1) with the standard topology which is complete.
In the early chapters we will develop several ways of producing a complete metric
from a given metric.
A space is said to be separable if it contains a countable dense subset. Again, the
space of real numbers is an example of a separable space because the set Q made
up of all rational numbers is countable and dense. A Polish space is a separable
topological space that can be metrized by means of a complete metric. From above
we see that the space of real numbers with the standard metric is an example of a
Polish space.
If X is a Polish space a subset A of X is analytic if there is a Polish space Z and
a continuous function f : Z ^ X such that f ( Z ) = A. The idea of the analytic set
was introduced by Souslin and Lusin and was developed mainly by the Polish and
Russian schools of mathematics between the First and Second World Wars. We will
use analytic sets to study measurable spaces. We will show that every Borel subset
of a Polish space is analytic and we will also prove the existence of a non-Borel
analytic set.
If (X, A) is a measurable space, a subset of X is said to be universally measurable
with respect to (X, A) if it is ^-measurable for every finite measure fx on (X, A).
In the final chapter after showing that each analytic subset of a Polish space X
is universally measurable with respect to (X ,B(X) ) where B(X) is the class of all
Borel subsets of X , we will be able to conclude that each Polish space contains a
non-Borel set which is universally measurable.
CHAPTER II
POLISH SPACES
2.1 Countability Axioms
Before we begin our study of Polish spaces we first need to define a few useful
topological properties. First recall that a space X is said to have a countable
basis at x if there is a countable collection B of neighborhoods of x such that each
neighborhood of x contains at least one of the elements of 3. A space that has a
countable basis at each of its points is said to satisfy the first countability axiom.
A topological space X is said to satisfy the second countability axiom if X has a
countable basis for its topology. Any set having the discrete topology or any set
which is metrizable would satisfy the first countability axiom. If the space X is
metrizable, an example of a countable basis at any point x £ X is the set that
consists of all balls centered at x with radius r, where r is a rational number.
Any countable set having the discrete topology would clearly satisfy the second
countability axiom. However an uncountable set having the discrete topology would
be an example of a space which satisfies the first but not the second countability
axiom.
The countability axioms are useful in the study of Polish spaces because of the
relationship between second countability and separability. In general, the property
of separability is a weaker property than second countability. The following re-
sult shows that any topological space which has a countable basis also contains a
countable dense subset.
Propos i t ion 2.1.1. If X is second countable then X is separable.
Proof. Suppose X is second countable. Let B = {B{ ji E N} be a countable basis
for X. For each i 6 N choose a point bi E Bi, and let S = [J{M- Then S is
countable because N is countable. Obviously S C X. Let x E X and U be an
open set containing x. Choose a positive integer j such that Bj C U and x E Bj.
Then bj E U and bj E S. Therefore every open set containing x intersects S and
x E S. Consequently, S = X. Since X contains a countable dense subset, X is
separable. •
Now we have that every second countable space is separable. However the con-
verse is not necessarily true. Consider the following example. Give the set of real
numbers R the lower limit topology. The set of all intervals of the form [a, b) where
a and b are real numbers form a basis for JR.. The set R with the above topology is
often called the Sorgenfrey line. Clearly, Q is a countable dense subset of R. Now
suppose B is a basis for R. Choose for each x E R an element Bx in B such that
x £ Bx C [x, x + 1).
If x then it is clear that Bx ± By because x = gib Bx and y = gib By. Conse-
quently B must be uncountable because R is uncountable. Therefore the Sorgenfrey
line is an example of a space which is separable but which is not second countable.
However using the following result we see that if a space is metrizable the properties
of separability and second countability are equivalent.
Propos i t ion 2.1.2. Let X be a metrizable space. If X is separable, then X is
second countable.
Proof. Let X be a separable, metrizable space. Let d be a metric for X, and let
S = {xi\i E N} be a countable dense subset of X. Let
B — {Bd(x i , r ) \x{ E S and r E Q}.
Then B is countable because S and Q are countable. Also B is a collection of open
sets in X with the metric topology.
Let U be an open set in X and let p G U. Because S is dense in X, p G S.
Case 1: p G S. Since U is open, q G Q can be found such that Bd(p,q) C U.
Therefore we have p G Bd(p,q) G B.
Case 2: p ^ S. Choose e > 0 so that Bd(p, e) C U. Such an e exists because U is
open. Also, Bd(p, | e ) is open. Choose j 6 N such that xj G S and xj G Bd(p, | e ) .
Since p S and xj G S, p ^ xj. Therefore 0 < d(p,xj) < | e . Choose rj G Q so
that
0 < d(p, Xj) <rj< ^e.
Hence p G Bd(xj,rj). Let y G Bd(xj,rj). Then
2 2 1 1
d(p,y) < d(p,Xj) + d(xj,y) < - e + - e = e.
Therefore p G Bd(xj,rj) C Bd(p, e) C U.
In both cases there exists an element of B containing p which is a subset of U.
Therefore B is a basis for the metric topology on X. Since B is countable, X is
second countable. •
We could conclude from the above argument that the Sorgenfrey line is not
metrizable. Consequently the Sorgenfrey line is an example of a space which is not
Polish. Now we have that each Polish space has a countable basis. The following
result is also useful.
Propos i t ion 2.1.3. If X is a second countable topological space and if Y is a
subspace of X, then Y is second countable.
Proof. Suppose X is second countable. Let B = {Bi\i £ N} be a countable basis
for X. Then C == {Bi Pi Y\Bi E B} is a countable basis for the subspace topology
on Y. Therefore Y is second countable. •
In the remainder of the paper if A is a subset of the topological space X then
Ac will be used to denote the complement of A in X . We can now use the above
theorems to establish the following theorem about Polish spaces.
Proposi t ion 2.1.4. Each closed subspace, and each open subspace of a Polish space
is Polish.
Proof. Let X be a Polish space and let Y be a subspace of X. Then X by definition
is separable and metrizable. Also, X is second countable by Proposition 2.1.2.
Therefore Y is second countable by Proposition 2.1.3. Consequently Y is separable
by Proposition 2.1.1. Let d be a complete metric for X.
Case 1: Suppose Y is closed. Since X is a Polish space and Y is a subspace of
X every Cauchy sequence in Y converges to a point in X. Since Y is closed every
Cauchy sequence in Y converges to a point in Y. Therefore, the metric inherited
by restricting d to Y is a complete metric for Y. Hence each closed subspace of X
is Polish.
Case 2: Suppose Y is open. Without loss of generality suppose that Y ^ X.
Recall that d(x,Yc) = i n f { d ( x , z ) : z £ K c } for all x £ X. Define the function
d0 : Y x Y —> R by
d0{x,y) = d(x,y) + 1
d(x,Yc) d(y,Yc)
We need to show that dQ is a metric for Y. Note that if x £ Y then d(x, Yc) > 0
since Y° is closed.
i) Since d(x,y) > 0 for all x,y £ X and absolute values are non-negative,
d0(x,y) > 0 for all x,y £ Y. Also, d(x,y) > 0 if x ^ y, thus dQ(x,y) > 0 if x ^ y. 1 1 If x = y, d(x, y) = 0 and
d(x,Yc) d(y,Yc) ii) Since d(x,y) = d(y,x),
do(x,y) — c?(a;,y) +
= 0 therefore d0(x, y) = 0.
d{x,Y°) d(y, Yc)
d(y,x) +
d0(y,x).
1
d(y,Yc) d(x, Yc)
iii) Now suppose that x,y,z £ Y. Then
d0(x,z) = d(x,z) +
< d(x,y) + d(y,z) +
= d0(x,y) + d0(y,z).
1
d(x,Yc) d(z, Yc)
1
d(x,Y°) d(y, Yc) +
d{y,Y°) d(z, Y c )
Therefore d0 is a metric in Y. Let Td0 be the topology inherited from dQ. The
next few paragraphs demonstrate that T(i0 is equivalent to the subspace topology
Td inherited from the metric topology on X.
Suppose that y £ Y and e > 0. Then B<i(y, e) is a basis element for Td containing
the element y. Consider Bda(y,e), which is a basis element for Tj0. Let x £
B<i0(y,e). Then d0(x,y) < e. It follows that d(x,y) < e. Hence x £ Bd(y,e).
Therefore, y £ Bda{y, e) C Bd(y, e) and Tda is finer than Td.
Again let y £ Y and e > 0. Then Bda (y, e) is a basis element for Tda containing y.
Since y EY and Y is open in X, there exists a neighborhood of y which is disjoint
from Yc, giving d(y, Yc) > 0. Choose S > 0 such that
4 < min{r- 2 S i '
Consider Bd{y,&). Let x £ Bd(y,S). The following argument shows that
d(x,y) > \d(y,Yc) - d(x,Yc)\.
Let z e Y c . Then
d(x,z) < d(x,y) + d(y,z)]
consequently,
inf{d(x, z) : z € Yc} < inf{d(x, y) + d(y, z) : z € F c } .
Equivalently,
and
Similarly,
thus
Hence
d(x,Yc) < d(x,y) + inf{d(y, z) : z £ Yc}
< d(x,y) + d{y,Y%
d(x,Yc) - d(y,Y°) < d{x,y).
d(y,z) < d(x,y) + d(x,z);
inf{d(y,z) : z 6 Yc} < inf{d(x,y) +d{x,z) : z 6 Yc}.
and
d{y,Yc) < d(x,y) + xni{d{x,z) : z e Yc}
< d(x,y) + d(x,Yc),
d(y,Yc) - <%,FC) < d(x,y).
Therefore \d(y,Yc) - d(x,Yc)\ < d(x,y).
Now it needs to be shown that x € Bdo(y, e). First we find that
d0(x,y) = d(x,y) -f
= d{x,y) +
d(x,Yc) d(y,Yc)
d(y,Yc) - d(x,Yc)
d(x,Yc)d(y,Yc)
< d(x,y) + d(x,y)
d{x,Y°)d(y,Y°)
Also, since
d(x,y) > |d(y,Yc) - d{x,Yc)\ > d(y,Yc) - d(x,Yc),
we have that
d(x,Yc) > d(y,Yc) - d(x,y) = \d(y,Yc) - d(x,y)|,
which yields the inequality
d0(x,y) < d(ar,j/) +
Since d{x, y) < 5,
d(x,t/)
d(x,y)
d(y,Yc)[d(y,Yc) - d(x,y)]
d(y,Yc)[d(y,Yc) - d(x,y)\ <
d(y,Yc)[d(y,Y<) - d(x,y)\
Moreover,
and
d(y,Y°) - d(x,y) > d(y,Yc) - S,
d(y,Y")[d(y,Y") - d(x,y)} > d(y,Y°)[d(y,Y°) - <S],
I d ( y , Y c ) \ d ( y , Y c ) - rf(x,!/)]| > \d(y,Yc)[d(y,Y<) - i] | .
Therefore,
d{y,Yc)[d(y,Yc) - d(x,y)]
Recall that S was chosen so that
< d{y,Y*)[d{y,Y*) - 5}
c\2
5 < ed(y,Yc)
2 + ed(y,Y°Y
10
Consequently,
2S + Std(y,Yc) < ed(y,Yc)2,
2S < Cd(y,Y')[d(y,Y') - <S],
and
> d(y,Yc)[d(y,Yc) - S\
Therefore, d0(x,y) < | e + \e = e. Hence x £ Bdo(y,e) and therefore x G
Bd(y,$) Q Bd0(y,e), and Td is finer t han Td0. Since Td C Td0 and Tj 0 C Td,
Td = Tdo and d0 metrizes the subspace topology on Y.
All t ha t is left to be shown is tha t d0 is a complete metric. Let (xn)n be a
Cauchy sequence under d0. Since d(xi,xj) < d0{xi,xj), the sequence (xn)n is a
Cauchy sequence under d and converges to a point x € X. Suppose x £ Y. Then
x E Yc and d(x, Yc) = 0. Let N £ N. Choose M such tha t if m, n > M then
d0(xm, xn) < 1. Choose q> M such tha t
d(xM,Yc) d(xq,x) <
Nd(xM,Y°) + 1
Since
we have tha t
Hence
d(xq,Yc) < d(xq,x) < d(xM,Yc),
1 1 <
d(xM,Yc) d(xq,Yc)
d0(xM,xq) = d(xM,xq) + 1
> 1
d(xM,Yc) d(xq,Y°)
1
d(xM, Yc) d(xq,Yc)
1 1
d(xq,Yc) d(xMiYc)
11
d(xq,x) CI(XM,YC)
Nd(xM,YC) + 1 _ 1 > D(XM,Yc) D(XM,Yc
= N.
This is a contradiction to d0(xM? xq) < 1- Consequently x £ Y, and therefore Y is
complete under d0 and Y is Polish. •
In the introduction we observed that the interval (0,1) with the standard metric
is not complete. However the above proof gives us a means of finding a metric which
metrizes (0,1) with the standard topology which is complete.
2.2 Disjoint Unions
If (X, T) and (Y, <S) are two topological spaces with X and Y disjoint, we define
a topology on the union Z = X U Y by taking as our open sets O C Z for which
OC\X eT and OC\Y 6 S. We call the space Z with this topology the disjoint union O
of X and Y and denote it by X U Y. If (X a ,Ta) is a n y indexed family of topological o
spaces with Xa Pi Xp = 0 for a 7̂ /3, we define their disjoint union Z = [jXa to
be the union of the Xa with O C Z defined to be open if O D Xa £ Ta for each
a. Each space Xa is referred to as a direct summand. If the sets Xa are not all o
disjoint, we take Ya = Xa x a and consider the disjoint union [jYa. The following
results show that many of the topological properties of a disjoint union depend only
on the topology of each of the direct summands.
o
Proposit ion 2.2.1. Let Z = {jXa. Then
a) the map f : Z —> Y is continuous if and only if each restriction f\xa i>s
continuous}
b) a set F C Z is closed if and only if F D Xa is closed for each a, and
c) Z is Hausdorff if and only if each of the spaces Xa is Hausdorff.
12
Proof, a) Suppose tha t / : Z —> Y is continuous. Let U be an open set in Y.
We assert that (f\xa D Xa for all a. First let x G ( / \ x a ) ~ 1 { U ) -
Then x G Xa and f{x) G U. Consequently, x G fl X a and ( / | x a ) - 1 ( £ 0 Q
/ - 1 ( ? 7 ) n X a . Now let X e f~l{U) n Xa. Then f{x) G U and x G Therefore
* £ ( / U J - 1 ^ ) a n d
( / k a r 1 ( ^ ) = r 1 ( ^ ) n x Q
as desired. Because / is continuous, / - 1 (Z7) is open in Z. Also, since Z is a disjoint
union / - 1 ( J 7 ) 0 is open in Z for each a. Therefore ( / | x a ) _ 1 ( ^ ) is open and
f\xa is continuous for each a.
Now suppose that f\xa is continuous for each a. Let U be open in Y. From
above, we have that ( f \ x a )_ 1 ( ^ ) = / _ 1 (U)C)Xa. Since f\xa i s continuous for each
a, we have tha t / - 1 ( { 7 ) D Xa is open for each a. Therefore by definition of disjoint
union, f~l(U) is open in Z. Hence / is continuous.
b)Let F C Z be closed. Then Fc is open and Fc fl Xa is open for each a. Since
Fc fl Xa = Xa — (Xa D F) for each a , Xa — (Xa fl F) is open in Xa for all a , and
consequently Xa D F is closed in Xa for all a.
Now suppose that Xa fl F is closed for each a. Then Xa — (Xa fl F) = Fc fl Xa
is open for each a. Therefore Fc is open in Z and F is closed in Z.
c) Suppose Z is Hausdorff. Let x and y be elements of Xa for some a. Since x
and y are also elements of Z there exist open sets U and V in Z such tha t x G U,
y G V and U fl V = 0. Because Z is a disjoint union, U Pi Xa and V D Xa are open
in Xa. Since x G U fl Xa, y G V D Xa and (U fl Xa) f l ( F n Xa) = 0, we have that
X a is Hausdorff for each a .
Now suppose for each a we have that Xa is Hausdorff. Let x,y G Z with x ^ y.
Then x G Xai and y G Xa. for some i,j. Suppose i =/=• j. Then we have that
Xai fl Xa. = 0 and both Xai and Xaj are open in Z. If i = j there exists disjoint
13
open sets U and V in Xa{ such that x <E U and y G V. Since U n Xp = 0 and
V H X/3 = 0 for (3 ̂ a i we have that U and V are open in Z. In either case we have
that x and y can be separated by open sets. Therefore Z is Hausdorff. •
Therefore, if a space is a disjoint union we can determine the properties of the
disjoint union by studying the properties of the individual summands. In the next
proposition we learn a little more about these direct summands.
P r o p o s i t i o n 2 .2 .2 .
a) If a subset Xi of a topological space X is a direct summand (in some disjoint
union), then Xx is both open and closed in X.
b) If is a subset of X which is both open and closed, then X = X\ U X2 where
X 2 = X - X 1 .
Proof, a) Suppose X = X\ UX2. Then, since X is open and X\ is a direct summand,
X fl Xi = Xi is open in X. Similarly, X2 is open in X . This implies that X - X2
is closed in X. Since Xx n X2 = 0 and X = Xx U X2 we have that X - X2 = Xx
and therefore X\ is also closed.
b) Suppose that Xx C X is both open and closed. Let X2 = X - Xx. Clearly
X = Xi U X2 and X\ H X2 = 0. Let 0 be an open set in X. Then we have that
Xx n O is open in X\ with the subspace topology and X2 fl O is open in X2 with
the subspace topology. Therefore 0 is open in Xx U X2. Now suppose 0 is an open
set in Xx U X2. Then 0 = (OflXi)U(On X2). Since O is open in Xx U X2, 0 fl Xx
and O D X2 are open in X] and X2 respectively. Since both Xx and X2 are both
open in X we have that 0 is open in X. Therefore X = Xx U X2. •
To obtain some of our results, it will often be necessary to construct a bounded
equivalent metric from a given metric. The following proposition gives us a method
for constucting such a metric.
14
Proposit ion 2.2.3. Let X be a metric space with metric d. Then
d0(x,y) = min{d!(x, y), 1}
is a bounded metric that induces the given metric topology of X, and X is complete
under d0 if and only if X is complete under d.
Proof. It first must be shown, that d0 is a metric.
i) Clearly da(x, y) — 0 if and only if d(x, y) = 0. Since d is a metric we have that
d0{x,y) = 0 if and only if x = y. If x ^ y then d0(x,y) = d(x,y) or d0(x,y) = 1.
In either case d0(x,y) > 0.
ii)If d0(x,y) = d(x,y) then d(x,y) = d(y,x) < 1. Therefore
do(y, x) = d(y, x) = d(x, y) = d0(x, y).
If d0(x,y) = 1, then d(y,x) = d(x,y) > 1. Therefore d0(y,x) = 1 = d0(x,y). In
either case d0(x,y) = d0(y,x).
iii)Suppose x, y, and z are three points in X. Either d0(x,y) + d0(y,z) < 1 or
d0(x,y) + d0(y,z) > 1. If d0(x,y) + d0(y,z) < 1 we have that d0(x,y) < 1 and
d0(y,z) < 1. This implies that d0(x,y) = d(x:y), d0(y,z) = d(y,z) and
d(x, z) < d(x, y) + d(y, z) = d0(x, y) + d0(y, z) < 1.
Therefore d0(x1zs) d^x^z^j d0(x^y^j c?0(y,z). If d0(y,z^ ^ 1 then
clearly d0(x,z) < d0(x,y) + d0(y, z).
Consequently dQ is a metric. Now let x 6 X and let e > 0. Let S = min{e, 1}.
Then clearly Bd0(x,5) C Bd(x,e). This implies that the topology on X inherited
from d0 is finer than the topology on X inherited from d. Again suppose that
x € X and e > 0. If e < 1, then Bd(x,e) = Bdo(x,e). If e > 1 then Bd(x, 1) C
X = ^d0{x,e). Consequently the topology on X inherited from d is finer than
15
the topology on X inherited from d0. Therefore the two metrics induce the same
topology. Since a sequence is Cauchy with the metric d if and only if the sequence
is Cauchy under the metric d0 we have that X is complete under the metric d if
and only if X is complete under the metric d0. •
In later proofs, we find that it is often useful to form disjoint unions from existing
Polish spaces. Above we have found that disjoint unions often preserve the proper-
ties of the individual summands. Now we find that if each of the direct summands
is Polish, then the disjoint union is also Polish.
Propos i t ion 2.2.4. The disjoint union of a finite or infinite sequence of Polish
spaces is Polish.
O
Proof. Let Xi,X2,X$,... be Polish spaces and let Z = [jXn be their disjoint union.
For each n let D„ be a countable dense subset in Xn and let dn be a complete
metric for Xn. Using Proposition 2.2.3 we can assume that dn(x, y) < 1 for each n
and each x, y G Xn. Define d such that for each x,y G Z dn(x, y) if x, y £ Xn for some n
d(x,y) 1 otherwise.
It is easily seen that d defines a metric on Z. We need to show that the topology
induced by d is the topology of Z. Let x G Z and e > 0. Then x G X j for some
positive j. Clearly if e > 1, Bd(x,e) = Z which is open in the disjoint union.
If e < 1, Bd(x,e) = Bdj(x,e) which again is clearly open in the disjoint union.
Similarly if U is open in the disjoint union, then there exists e such that 0 < e < 1
and Bdj(x,e) C U. Consequently Bj(x, e) C U. Therefore d metrizes the topology
of Z. Let be a Cauchy sequence in Z. For some N G N if m , n > N we have
that d(xm,xn) < 1. Therefore for all i > TV, Xi G X j for some j. This implies that
(xi)i^zN a Cauchy sequence in X j and therefore converges to a point of X j . Since
X j C Z, (xi)i converges to a point in Z. Now we have that Z is complete.
16
All that remains is to show that D = | J n Dn is a countable dense subset of Z.
Clearly since each of Dn is countable and D is the countable union of the sets Dn.
D is also countable. Let U 7̂ 0 be an open set in Z. Then for all i £ N, U D Xi is
open in X{. For some some j we have that U fl X j 7̂ 0. Since Dj is dense in X j
there exists an x £ U fl X j such that x £ Dj. Since Dj C D we have that x £ D.
Also since x £ U fl X j we have that x £ U. Therefore U contains a point of D and
D is dense in Z. Since Z is metrized by a complete metric and contains a countable
dense subset, Z is Polish. •
2.3 Product Spaces
Now we turn our attention to product spaces. Recall that if (X, 7~) and (7, <?)
are two topological spaces, we define a topology on the product 1 x 7 by taking
as a base the collection of all sets of the form U\ x U2 where U\ £ T and U2 £ <S.
This is called the product topology for X x Y. If (XaiTa)a is any indexed family of
topological spaces, we define the product topology on ]~|a Xa by taking as a base all
sets of the form aUa, where Ua £ Ta and Ua = Xa except for a finite number of
a . In the following proposition another method of constructing a bounded metric
is used. We see that if p is a metric for the space X then the metric p* given by
p* = is a bounded metric on X which yields the same topology as p.
Propos i t ion 2.3.1. Let (Xk,Pk)k be a sequence of metric spaces. We define their GO
direct product Z = X& by letting Z be the space of all sequences (xk)k with k=1
CO Xk G Xk and defining a metric r on Z by setting r(x,y) = ^ ^~kP\{xk-> Vk) where
k=i X = (xk)k, 'y = (yk)k and pi = Then
a) r is a metric and the sequence in Z where X{ — {xn, Xi2,...) converges to
y E Z if and only if (xij)(^zl converges to yj for each j,
b) (Z7t) is complete if each (Xk,PK)
17
are c) and if for each k the space (X\k,Pk) and {Yk,&k) are homeomorphic, then so oo oo
the spaces n xk and n Yk. k=1 k=1
Proof, a) Let Z and r be defined as above. Let x,y E Z. For each i E N,
pi{xj,yi)
1 + pi(xi,yi) < 1 therefore
OO / \ oo
r(x,y) = < y 2 - ' = i.
Consequently, r : Z x Z —>• R.
i) Again let E Z. Since pk(xk,yk) > 0 for all k, r(x,y) > 0. Since pk is
a metric for each k1 if x = y, xk = yk for each fc, and pk(xk,yk) = 0 for each k.
Therefore
r ( x , x ) = f ; 2 - t1
/ " ' ( l i ' X t ) , = 0 . fc_ - + Pk{xki Xk)
Now suppose r(x,y) = 0. Then
2_ f c pk(xk,yk) _ Q
1 + Pk{xk,yk)
for all k. Thus pk(xk,yk) = 0 , giving xk = yk and consequently x = y as desired.
ii)Clearly r(x,y) = r(y,x) because pk(xk,yk) — Pk(yk,xk) for all k.
iii)All that is let to show is that r satisfies the triangle inequality. Let x,y,z E Z,
then
pk(xk,yk) . Pk(yk,zk) r(X,y) + r{y,z) = £ £ 2" l + w ( w , , t )
~kPk{xk,yk)[l + pk(yk,Zk)] + pk(yk,Zk)[l + p(xk,Vk)\ = £ 2 ~
k=1 (1 + pk(xk,yk))(i + pk(yk,zk))
°° -k Pk{xk,yk)+ 2pk(xk,yk)pk(yk,zk) + Pk{yk,zk) E 2 " * t k= 1
+ pk(xk,yk)pk(yk,zk) + Pk(xk,yk) + Pk(yk,zk)
> ^ n - k pk(xk,yk) + 2pk{xk,yk)pk{yk,zk) + pk(yk,zk) E 2 " ' r k= 1
+ 2pk(xk,yk)pk(yk,zk) + pk(xk,yk) + pk(yk,zk)
18
= T . 2 - " — , — k=i pk(xkiVk j+i pk(%k ,yk) pk(yk >zk)+pk(yk ,zk) oo -
> E 2 - ' , ' , + 1
k= 1 pk(Xk^k) Pk{xk^k)
k=1 l + Pk(xk,Zk)
= r(x7z).
Hence r is a metric.
Let (xi)i be a sequence in Z which converges to y, where Xi = (xn, Xii,...) for
each i and y = (yi , j/25 L ( ,t j 6 N and let e > 0. F ind 8 > 0 such tha t 8 < ,
giving tha t 8 8e < e and Choose N <E N such tha t n > iV implies tha t
r ( « n , y ) < Then
, . _ 0 - f c PkjXnkiVk)
r { x n , y ) - ^ l l + p ^ . y f c ) < 2> /C= 1
for all n > N. This implies t ha t if n > N, then
2 -i Pj(Xnj->yj) ^ 1 + pj{xnj,yj) 2-?'
P j ( X n j , V j ) ^ ^
1 + pj(xnj,yj)
P j { x n j 1 V j ) < $ + ^Pj{xnj') Vj)i
Pj(xnj,yj)( 1 -5) < ^
(J Pj{Xnj->Vj) <
1 - < T
and
P j ( x n j i V j ) < e*
Hence converges to y j for all j G N.
Now suppose (#m)n?=i converges to yj for all i £ N. Let e > 0 and choose N <E N
such tha t for all n > TV oo 1
£ 2 - ' < r . k=n-\-1
19
For each i < N, Mi can be found such tha t pi(xni,yi) < for all n > Mi. Let
M = max{iV, Mi , M 2 , M j v } - Let n > M. Then
w / t ( X . , , , ) = £ 2
1 +pi(a ;ni ,y i ) 2=1 N
Pifanii Vi)
4" Pii^xni^ y>t
__ o~?' ^ g - t
j=1 1 Pi(xnif Hi) i—N+i ^ P^Xn^V 0
N oo / x
^ / - >k i o - i Pi\xni,yi)
;=»+! 1 + * (*« . , ! / , )
N oo
^ E 2 _ i | + E 2 _ i
i-1 i = i V + l
^ 1 1
~ 2 2
= e.
Therefore (arn)ra converges to y in Z .
b)Let {xi)i be a Cauchy sequence in Z. Let e > 0 and let j £ N. Choose N such
tha t r{xn, xm) < 2~- 7 (^y) for all n , m > N. Then for each n,m > N we have tha t
E0 - t W t \ „ 0 - j 6
2 — _ = r (a ; B ) a ; m ) < 2 J — — j- ^ 1 "t~ Pi\Xni 5 ^ 1
This implies tha t
2~i Pj(xnj) amj) ^ 2~j e
1 -)- P j{xnj ^Xmj^ 6 1
and therefore P j ( x n j , x m j ) < e. Consequently is a Cauchy sequence in
Xj for all j. By hypothesis ( x n j ) 1 converges to the point yj and by par t (a)
this implies tha t (x,), converges to the point y = (y1,y2,...) G Z. Therefore Z is
complete.
c) Assume tha t for each k the spaces (Xk, pk) and (Yk, crfc) are homeomorphic. For O O
all A; let fk : Xk —> Yk be a homeomorphism from Xk onto Yk. Define / : Xk —> k=i
20
oo
n Yk by f(x) = (fi(xi), f2(x2), fs{x3)...) where x = (xi,x2,x$,...). Since each fk k=1 is a bijection, it is clear that / is also a bijection. Suppose (xi)(-Z1 is a convergent
oo sequence in Yl Xk• By part (a), x is a convergent sequence in Xj for each j.
k=1 Since f j is a homeomorphism we have that is a convergent sequence in
oo Yj. Again by part (a) we have that ( f (x i ) ) f l 1 is a convergent sequence in Yk.
k=1 Therefore / is continuous. The proof that / _ 1 is continuous is similar. Therefore
oo oo oo the spaces -^k and ]~| Yk are homemorphic. The proof is similar that if n xt
k= 1 fc=l k= 1 oo
and Yk are homeomorphic then for each k the spaces (Xk,pk) and (Yk,crk) are k= 1
also homeomorphic. •
Proposit ion 2.3.2. If (X,p) and (F, a) are two metric spaces then the product
topology on X x Y is the same as the topology induced by the product metric.
Proof. Let T represent the product topology on X x Y, define r on X x Y by
r({x1,yl),(x2,y2)) = \J p{x i,x2)2 + o"(yi,y2)
2,
and let S represent the topology induced by the metric r . Let (a, b) 6 l x F and let
A x 5 b e a basis element of (X x Y, T) which contains (a, 6). By definition of product
topology, A is open in X and B is open in Y. Consequently there exist > 0 and
eg > 0 such that Bp(a, e^) C A and Ba(b, eg) Q B. Let e = m i n l e ^ e s } ; then
(a, b) £ Bp(a, e) x Ba(b, e) C A x B. Let (c, d) £ Br((a, 6), e); then
\/V(a,c)2 +<j(6, d)2 = r((a, 6), (c, d)) < e.
This implies that p(a, c) < e and <r(6, d) < e. Therefore (c, G?) £ A X B and
Br((a, b), e) C 4 x 5 . Hence TCS.
Now let (a,b) e X x Y and let e > 0. Let S = ^ and let (c,d) € Bp(a,S) x
5^(6,5). Then
-((a,6),(c,d)) = yjp{a,c)2 + <t(6,o?)5
21
< v ^ 2 + £2
" V ¥ + 2 "
= e.
Hence (c,d) G Bp(a,8) x B(r(b,S) C _Br((a, 6), e). Consequently S CT. Therefore
the two topologies are equal. •
Now we return to our study of Polish spaces.
Proposit ion 2.3.3. The product of a finite or infinite sequence of Polish spaces is
Polish.
Proof. Let Xi,X2,X$,... be a finite or infinite sequence of Polish spaces. Assume
that Xi ^ 0 for all i. For each n let dn be a complete metric in Xn such that
dn(x,y) < 1 for all x,y G Xn. Also define d such that
d{X) y) = ^ ^ 2 d n (x n , y n ^ n
where x = (xux2,...) and y = (t/i,t/2, •••)• Let X->V € \[Xn. n
i) If d(x,y) = 0 it must be that dn(xn,yn) = 0 for all n. Since each dn is a
metric, it must be that xn = yn for all n which implies that x = y. Also if x = y it
is obviously true that d(x,y) = 0.
ii) Clearly since dn(xn,yn) = dn(ynixn) for all n we have that d(x,y) = d(y,x).
iii) Let x,y,z G \\Xn. Then n
y) -|- z) ^ ^ 2 dn[xn^ yn) ^ ^ 2 dn{yn^ zn) n n
^ ^ 2 [dn[xn^ yn) 4~ dn(yn^ zn)~\ n
^ 2 dn[xn^zn) n
= d(x, z).
22
Therefore d is a metric for the space JJ Xn. n
Next we show that the topology Td inherited from d is the same as the product
topology, Tp. Let x £ J ] Xn. Let U be a basic open set in Xn, Tp) which contains n n
x. Then
U = Ui X U2 X U3 x ... X Ui x Xi+1 x Xi+2...
for some choice of i £ N and U& open in Xk, k < i. For each Uk choose an such
that 0 < €k < 1 and Bdk(xk, efc) Q Uk• Let e = minje i , e2, e3,..., e;}, and let
V = Bdl(xi,e) x Bd2(x2,e) x ... x Bd{(xi,e) x x X i + 2 x ....
Then V C U. Consider Bd(x, 2 V) which is open in X n with the topology Td. n
Let y £ Bd{x, 2~le). Then
d(xj t/) — ̂ ^ 2 j/n) 2 e. n
Consequently for all n < i,
2 dn(^n, ) < 2 €
and dn(^n5J/n) < < e. Therefore y G V C £7, 2~~*e) C 17, and C T^.
Again let x G J ] and let 0 < e < 1. Note that for all x, y G f ] we have n n
d(x,y) = < 1. n n
Consider Bd(x, e) which is a basis element for Td which contains x. Since the sum
^ 2 - r a < oo we can choose N such that 2~n < | e . Let « n>iV
Z7 = 2^:) X Bd2(x2, X ... X BdN(xN, ^~r) X Xjy+i X ...
and let y £ {/. Then
?/) ^ ^2 dn(xn,yn^
23
^ ^ 2 dn[xn,yn^ -f- ^ ^ 2 dn(xn, y n ) n<iV n>N
^ ^ ^ (^n ? Un ) H~ ^ ^ 2 dn (xn, yn) n<.N n>iV
^ Ne 1 < |_ —6 ~ 2 i V 2
= 6.
Therefore {7 C e), C Tp , and we have that d metrizes the product topology
on l\Xn. n
All that is let to show is that the metric d is complete and that the product space
is separable. Let ( x n ) n be a Cauchy sequence under <i, where for each n, the point
%n = (^ni j xn2-> Xn3,...). Fix j and let e > 0. Consider the sequence in X j .
Choose N G N such that m , n > N implies that d(xn,xm) < 2-«?e. Then
d^XfijXffi^j — ̂ ^ 2 dk{^Xjik, ) <C 2 ^6. k
Consequently
2 ^d j (x T l j , xmj) <C 2 *̂ 6,
and
dj , %mj ) "\ ۥ
Since this is t rue for all m, n > N we have that is a Cauchy sequence in X j .
Since X j is Polish, the sequence converges to a point of X j . Since this is true for all
J5 the sequence ( z n ) n converges to a point of \\Xn. Therefore ]\Xn is complete n n
under d.
For each n choose a countable base Bn for Xn. Let
$ — {Ui x U2 x ... x Un x Xjv+i x ...\Ui G Bi and N £ N}.
24
Then B is a base for J~[ Xn and B is countable. Since ]^| Xn is a metrizable space with n n
a countable base, by a previous theorem Xn is separable and therefore Polish. • n
Now we have that the product of a sequence of Polish spaces is Polish. The next
result gives a rather explicit way of obtaining a countable dense subset.
L e m m a 2 ,3 .4 , Suppose that (Xnjdn)n is a sequence of non-empty separable met-
rizable spaces and that
Dn —• ^ G
is a countable dense subset of Xn for each n. If an is chosen arbitrarily in Xn for
each n G N; then the set
D = {(xi i 1 ?x 2 i 2 , ...,XiVijV,ajv+i,aiv+2? •••) : N £ G Dj for 1 < j < N}
is a countable dense subset ofY[Xn. n
Proof. By the proof of Proposition 2.1.2 the set
Bn = {Bdn(xni,r)\xni G Dn and r G Q}
is a countable base for (Xn,dn). Let U be the collection of sets of the form
Bd1{xul,rl) x Bd2(x2i27r2) x ... x BdN (xNiN, rN) x XN+t x XN+2 x ...,
for some choice xjii G Dj and r ? G Q, and for some N G N. The set U is a countable
base for n nXn. For each Xn choose an element an G Xn. Let D be the set of
elements of the form
for some choice xjij G Dj and N G N. Then D has a one-to-one relationship with
U. Consequently D is countable.
25
Let A be an open set in J}n Xn. Then there exists a set
B = Bd1 (x i^ , r i ) x Bd2(x2i2,r2) x ... x BdN(xNiN,rN) x XN+1 x XN+2 x
which is a subset of A. Then
Hi 5 ®Nijv ? aiV+2 j ^ ^
Therefore D is dense. •
2.4 Each Polish Subspace is a
We now have that open subsets and closed subsets of Polish spaces are Polish
and that disjoint unions and products of Polish spaces are Polish. The purpose of
this section is to show that a subspace of a Polish space X is Polish if and only if it
is a G$ in X. Recall that a subset Y of X is a Gs in X if Y can be written in the
form Y = P|n Un where each Un is open in X. Also, a subset Y of X is an in X
if Y can be written in the form Y = \JnVn where each Vn is closed in X. Before
we can show that a subspace of a Polish space X is Polish if and only if it is a Gs
in X, we first need the following results.
Proposition 2.4.1. Each closed subset of a metric space is a Gs-
Proof. Let C be a non-empty closed subset of a metric space X. For all n G N, let
1 1 An = jx € X
If x G C,
d(x, C) < n
d(x, C) = inf{<i(x, z)\z G C} = 0 < —,
n
and x G An. Since this was not dependent on the choice of n we have C C P|n An.
Now suppose x G p|n An. Then d(x, C) < ^ for all n. This implies that each open
set containing x contains a point of C and therefore x G C = C. Consequently,
26
n „ A , C C. Since we have containment in both directions, p | n An = C. We finish
the proof by showing that for each n the set An is open. Fix n and let x G An.
Then d(x,C) < Let 7 = d(x,C) and let 8 < ^ — 7. Note that 8 > 0. Let
y G B(x, 5); then
d(y, C) = in f{d (y , z)\z G C}
< in f {d (x , y) + d(x, z)\z G C}
= d(x,y) + inf{d(x, z)\z G C}
< 8 + 7
1 < n
Therefore y G An. Hence for each x G An an open neighborhood of x can be found
which is a subset of An. Therefore An is open for all n. Since C = f ] n An, the set
C is a Gs. •
T h e o r e m 2.4 .2 (Cantor's N e s t e d Set Theorem) . Let X be a complete metric
space. If (An)n is a decreasing sequence of non-empty closed subsets of X such that
limn diam(An) = 0; then P l n = = i An contains exactly one point.
Proof. Let X be a complete metric space. Let (An)n be a decreasing sequence of
non-empty closed subsets of X such that limn diam(j4n) = 0 . For each n G N
choose xn G An, and let e > 0. Since limn diam( J4n) = 0, a positive integer N can
be found such that diam(An) < e for all n > N. Let m,n > N. Since (An)n is a
decreasing sequence, xm,xn G Ajv; therefore
d(xm,xn) < diam(Aiv) < e.
This implies that (xn)^L1 is a Cauchy sequence. Because X is complete, (a;n)^i_1
converges to x G X. Let i G N. Then also converges to x. Since (An)n
27
is a decreasing sequence, each element of [xn)'^>
=i is an element of At which is
closed. Therefore x £ A{ for all i, and x £ HnLi An. Since limn diam(An) = 0,
if y £ 0^=1 it must be that x = y. Therefore An contains exactly one
point. •
Lemma 2.4.3. Suppose (X,d) is a metric space and that {An)n is a sequence of
sets such that limn diam(An) = 0 . Then limn diam(An) = 0.
Proof. Suppose that the hypotheses are satisfied. Let e > 0. Choose TV £ N such
that diam(Ara) < | for all n > TV. Let n> TV and x,y £ An. Since Bd(x, | ) is an
open set containing x we can choose w £ An such that w £ Bd(x, | ) . Similarly we
can choose z £ An such that z £ Bd(y, f )• Then
d{x, y) < d(x, w) + d(w, z) + d(z, y) < e.
Hence diam(An) < e. Since e was arbitrary limn diam(Are) = 0 . •
The following lemma is important in this section, but it will also be used often
throughout the remainder of this paper.
Lemma 2.4.4. Let (X, T) and (V, S) be topological spaces and let f •. X Y be a
homeomorphism. If (V, 5*) is a Polish space then (X, T) is also a Polish space.
Proof. Suppose the hypotheses are satisfied. Let d be a complete metric for (Y, S).
Define the function a : X x X —»• K. by
<r(x,y) = d(f(x),f(y)).
It needs to be shown that a is a metric for X. Let x,y,z £ X.
i) Since d is a metric d(f(x),f(y)) > 0; consequently, a(x,y) > 0 . If x — y,
f(x) = f ( y ) which implies that
a (x,y) = d(f(x)J(y)) = 0.
29
implies that ( / ( # i ) ) ^ i is a Cauchy sequence in Y. Since Y is complete
converges to some point y £ Y. Since / is surjective y = f(x) for some x £ X.
Again let e > 0. Choose N £ N such that d ( / (#n) , / (# ) ) < e for all n> N. Then
a (xn,x) = d(/(®«),/(s)) <
for all n > N. Hence converges to x £ X. Consequently cr is a complete
metric.
Let 5 be a countable dense subset of Y. Consider the set f~1(S) C X. Clearly,
since / is one-to-one, / - 1 (S) is countable. Let U be an open set in X and note that
f(U) is an open set in Y. Choose s € S such that s £ f(U). Then / - 1 ( s ) £ U.
Since is also in f~1(S) we have that / _ 1 ( 5 ' ) is dense in X. Since X is
separable and cr is a complete metric for X, we have that X is Polish. •
If Y is a subspace of X and A is a subset of Y it is possible that the closure of
A in Y is a proper subset of the closure of A in X. In the following proof A is used
Y
to denote the closure of A in X and A is used to denote the closure of A in Y.
Now we are ready to prove the main result of this section.
Proposit ion 2.4.5. A subset of a Polish space X is Polish if and only if it is a
G$ in X.
Proof. Let X be a Polish space and suppose that Y is a Gj-subset of X. Let (Un)n
be a sequence of open sets in X such that Y = (\nUn. By Proposition 2.1.4, Un
is Polish for all n € N. Also, by Proposition 2.3.3, Y\Un is Polish. Let A be the n
subset of IJ Un defined by n
A = < u = (ui , u2 , . . .) G Un Uj = u/c for all j,k >. n ^
Suppose x = (x\, #2,-••) is a limit point of A. For each i £ N choose U{ £ A
such that U{ = (un,Ui2,...) and the sequence ( u j ) ^ 1 converges to x. Note that for
30
all j G N, (uij)^ converges to xj. Since m G A, mj = uik for all j,k. Hence
x j — %k for all j,k and x G A. Consequently, A contains all of its limit points and
is therefore closed. By Proposition 2.1.4, A is Polish.
Let f : Y —y A be such that f ( y ) = (y,y,y,. . .). Note that since Y = f]nUn,
f ( y ) e A for all y e Y. Similarly if (x,x,x,...) e A, x € Ui for all i, giving x € Y.
L e t (yi)U 1 b e convergent sequence in Y. Since /(?/;) = (un, ui2, «i3, •••) where
uij = Vi f ° r a h ji we have ( ^ i = and therefore converges for all
j. Thus (f(yi))il1 converges. Similarly if (f(yi))il1 is a convergent sequence, the
sequence also converges. Because / and / - 1 are both continuous and / is
one-to-one and onto, / is a homeomorphism of Y onto A. By Lemma 2.4.4, Y is
Polish.
Now suppose the subspace Y in X is Polish. Let d be a complete metric for the
topology of X and let da be a complete metric for the topology of Y. For each n £ N
let Vn be the union of those open subsets W of X for which W D Y is non-empty and
WHY has a diameter of at most J under d0. Let us show that Y = Y D ( f | n K) -
First suppose y £ Y. Obviously y (E Y. Let n £ N. Then Bd0(jji ~) is open in Y
and is therefore equal to W fl Y for some W open in X. Since W f)Y = Bd0(y, -)
has a diameter of at most ^ in Y, W C Vn. Also since y 6 Bdo(y, ^) , we have
yew CVn. This is true for all such n, hence y E f | n vn a n d Y C Y n ( f ) n Vn).
Now suppose x € Y fl ( f | n Vn). This implies that x e Vn for all n. Since Vn
is the union of open sets, for each n choose Wn open in X such that x G Wn and
diamd0(Wn fl Y) < Notice that because x is also an element of Y, Wn Pi Y / 0.
By replacing Wn with a smaller open neighborhood of x we can assume without loss
of generality that is a decreasing sequence and that diamd{W n) < By
Lemma 2.4.3, we have that limn[diam(/o(VFra D Y )] = 0. Also since n was arbitrary
limn[diam(i(l/Fn)] = 0. Since Y is complete under d0 the Cantor Nested Set Theorem
31
implies that y E F can be chosen such that y E f)n{Wn fl Y Y ) . Since
wnc\Y c wn n F c wn n F
for all n, we have y E P|n Wn. Also x E Wn C Q n Wn. Therefore again applying
the Cantor Nested Set Theorem x = y. From above we have that y E F , therefore
x E Y. This gives that Y D (p|n Vn) C Y. Consequently Y = Y fl (P|n Vn). By
Proposition 2.4.1, F is a Gj , therefore Y is a G&. •
The Cantor set C is the set which consists of all those real numbers of [0,1]
that have ternary expansion (anJn for which an is never 1. If x has two ternary
expansions, we put x in the Cantor set if one of the expansions has no term equal
to one. The set C can be obtained by first removing the middle third (§, §) from
[0,1], then removing the middle thirds ( | , | ) and ( | , | ) of the remaining intervals,
and inductively continuing this process. Since the complement of the Cantor set is
the union of open intervals, the Cantor set is closed. This implies that the Cantor
set is a Gs in [0,1] (Proposition 2.4.1). Since the set [0,1] is clearly a Polish space,
by the above proposition the Cantor set is also Polish. The following proposition
gives us another way of representing the Cantor set that will be useful later in the
study of analytic sets.
Proposit ion 2.4.6. The Cantor set is homeomorphic to the set {0,1}N .
Proof. Define / : C -» {0,1}N by = (&;), where bi = 0 if a{ = 0 and
bi = 1 if a, = 2. Suppose (cn)n / (dn)n then for some i we have a ^ di which
implies / ( ( c n ) n ) ^ f((dn)n). Therefore / is one-to-one. Now suppose (bn)n E
{0,1}N. Define (an)n such that an = bn if bn = 0 and an = 2 if bn = 1. Then
/((«n)n) = (bn)n and / is onto. Let ( a ; ) -^ be a convergent sequence in the Cantor
set, where cii = (ajj, a^, . . . ) . Then, for each j , ( a ^ ) ^ 1 is a convergent sequence.
This implies that is eventually constant. Let (a{j )^ 1 converge to x j for
32
each j. Then clearly ( / ( a ; ) ?^ converges to f(x), where x = (xj)j. Therefore /
is continuous. Since C is compact and {0,1}^ is Hausdorff we have that / is a
homeomorphism. •
Another approach to finding a homeomorphism from {0,1}N to the Cantor set
is to view the Cantor set as the set of all elements of the form where each
rik takes on the value zero or one. Then map each element (rik)k in {0,1}N to
Efc n h - T o see that this map is one-to-one suppose that (Tik)k 7̂ ( m k ) k and choose
the smallest integer i such that m Without loss of generality suppose that
rii = 1 and m4- = 0. Then
0 0 0 0
rik ~ mk nk — mk Erik ~ m k _ y -
zk ~ ^ W k—1 k—i -j CXJ
h+ ? k=i-\-
1 00
OO
nk - m k
3k k=i+l
Therefore OO
°° 1 > - - V —
3« ^ 3* k=i-\-1
- 1 -1(1. ~ 3 i ~ 2
> 0.
2mk 3fc 2-J 3k •
k=1 k= 1
Therefore the map is one-to-one. The proof that this map is also continuous and
onto is similar to the proof above.
By Lemma 2.4.4 we can now conclude that the set {0,1}N is a Polish space.
The space N is another example of a Polish space because it is a closed subset of
the Polish space R. By Proposition 2.3.3, is also a Polish space. This space is
extremely useful in the study of analytic sets and will often be denoted by Af. Also
the space X of irrational numbers in the interval (0,1), together with the topology
33
it inherits from JR., is Polish because it is a Gg in (0,1). Now let us use Proposition
2.4.5 to show that Q is not Polish.
Proposition 2.4.7. The set Q of rational numbers with the topology it inherits as
a subspace of R is not Polish.
Proof Suppose that { x is a sequence of open subsets of R such that Q =
C\nUn. Then R — Q = R — C\nUn = Un(^ — Un). This implies that the irrationals
are an T^. Clearly for each n the set R — Un has an empty interior because every
open set would contain a point of Q and consequently would not be a subset of
R — Un. Since <Q> is countable and therefore also the union of nowhere dense sets
we have that R is the union of nowhere dense sets. This contradicts the Baire
Category Theorem. Consequently Q is not a Gs• By Proposition 2.4.5, Q is not a
Polish space. •
2.5 Borel Subsets of Polish Spaces
and Measurable Spaces
We turn to some basic facts about the Borel subsets of Polish spaces. First we
need a few definitions. A collection A of subsets of X is called an algebra if X £ A
and
i) A U B G A whenever A, B € A,
ii) A° 6 A if A 6 A,
Hi) A n B g A if A, B e A.
In fact if a collection A of sets satisfies (i) and (ii) then De Morgan's law would
imply that it also satisfies (iii). An algebra A of sets is called a a-algebra if every
union of a countable subcollection of sets in A is in A. Again using De Morgan's
law this would also be true if every intersection of a countable subcollection of sets
34
in A is in A. The collection B of Borel sets in a set X is the smallest u-algebra
which contains all the open sets of X. This set is often denoted B(X).
Let (Xi,A\) and (X2 , ,42) be measurable spaces. Recall that a function / :
X1 ->• X2 is said to be measurable with respect to A\ and A2 if for each B in A2 the
se^ f {B) belongs to A\. A function f '• X —y Y is said to be Borel measurable if
for each Borel set B in Y the set is a Borel set in X.
Let (X1,Ai), (X2,A2),... be measurable spaces. The product of these measur-
able spaces is the measurable space ( f j n Xn, An) where An is the <r-algebra
on n n X n generated by the sets that have the form
A\ x A2 x ... x An x Xjv+i x ...
for some positive integer N and some choice of An € An, when n = 1,2,... , JV. In
order to show that the class of Borel sets in a measure space has certain properties,
it is often useful to show that the sets which have those properties are a er-algebra
which contain the open sets. Since the collection of all Borel sets is the smallest such
er algebra, this would give us that the class of Borel sets also has those properties.
The following proposition gives us a better representation of the class of Borel sets
in product spaces.
Propos i t ion 2.5.1. Let Xi,X2,Xz,... be a finite or infinite sequence of separable
metrizable spaces. Then B{Y[n Xn) = J]n B{Xn).
Proof. By definition \[nB{Xn) is the (T-algebra on ]JnXn generated by the sets
that have the form
x B2 x ... x JBJV x A'Jv+1 x Xn+2 X ...
for some positive integer N and some choice of Bn £ B(Xn). Recall that the
projection function ^ : Y [ nX
n ^ X i defined by
~ CLi
35
is continuous. If A is a Borel set in Xi then
7T- 1(A) = Xi x X2 X . . . X X i - i X A x X i i+1 x •••
which is an element of Y\nB{Xn). Hence iri is measurable with respect to B ( X i )
and B { X [ n ( X n ) ) for each i. Note that if N € N and Bi G Xi for i < N then
N
Bi x B2 x . . . x Bn x Xn+i... = 7 r t- 1 ( J 5 i ) ;
i=i
therefore B ( X n j is the smallest a-algebra on Xn that makes each projection
7Vi measurable. The continuity of irl for each i implies tha t if A is open (closed)
in Xi, then (A) is open (closed) in X . Thus 7 1 ( A ) £ B ( Y l n X n ) . Since in-
verses preserve unions, for any Borel set B in X i , t t ~ 1 { B ) e B ( ] J n X n ) . Hence nt
is measurable with respect to $ ( r i n - X n ) and B(Xi). Since B(Xn) is the small-
est cr-algebra on J | „ X n tha t makes these projections measurable, it follows tha t
U n B ( X n ) C B ( U n X n ) .
For each n choose a countable base Un for Xn. Let U be the collection of sets
tha t have the form
U\ X U2 X . . . X UN X XN + i X Xp*j+ 2 X . . .
for some positive integer N and some choice Un <G Un. Then U is a countable
base for Y l n X n and U C Y [ n B ( X n ) . Since each open set of X [ n X n is a union of
elements of U and U is countable it follows tha t B ( \ \ n X n ) C [ \ n B ( X n ) . Thus
^ (11 n^-n) = \ [ n B { X n ) as desired. •
If X and Y are sets and / is a function from X to Y then the graph of / (denoted
by g r ( / ) ) is defined by
g r ( f ) = { ( x , y ) e X x Y : y = f ( x ) } .
The following results deal with measurable functions.
36
Proposit ion 2.5.2. Let X and Y be separable metrizable spaces and let f : X —>• Y
be Borel measurable. Then the graph of f is a Borel subset of X xY.
Proof. Let F : X x Y -4 Y x Y be the map that takes (x, y) to ( f ( x ) , y). The Borel
measurability of / implies that if A,B £ B(Y) then / _ 1 (A) and f~l ( B ) are in B(X).
This implies that F 1(A x B) G B(X) x B(Y)~, hence F is measurable with respect
to B(X) x B(Y) and B(Y) x B{Y). Therefore by Proposition 2.5.1, F is measurable
with respect to B(X x Y) and B(Y x Y). Let A = {(yx,y2) G Y x Y\yi = y2}. By
the proof of Proposition 2.4.5, A is a closed subset of F x Y .
We now need to show that gr( / ) = F ^ A ) . Let (x,y) 6 gr( / ) . By definition
f{x) = V- This implies that F((x,y)) = (f(x),y) g A. Consequently, gr( / ) C
^ " ^ A ) . Now suppose that (u,v) G F _ 1 ( A ) . Then F((u,v)) = (/(«), v) e A.
Hence f{u) — v, (u,v) G gr(/) , and gr( / ) = i ^ - 1 (A) . Since i*1 is Borel measurable
it follows that gr( / ) is a Borel set in I x F . •
Proposi t ion 2.5.3. Let (X,A), (Y,B ) , and (Z,C) be measurable spaces and let f :
('Y,B) (Z,C) andg : (X,A) (F ,0 ) 6e measurable. Then fog : (X , ^ l ) (Z,C)
is also measurable.
Proof. Suppose that C G C. Then / - 1 ( C ) G B and <7 - 1 ( / - 1 (C)) G «4. Since
( f 0 9) 1 (C) = S , - 1 ( / _ 1 (C)), the measurability of / o g follows. •
Let X be a set, and let J- be a family of subsets of X. Then the smallest <x-
algebra on X that includes T is clearly unique; it is called the cr-algebra generated
by T , and is often denoted by o{T).
Proposit ion 2.5.4. Let (X,A) and (Y,B) be measurable spaces, and let B0 be
a collection of subsets of Y such that <r(B0) = B. Then a function f : X Y
ts measurable with respect to A and B if and only if f~l{B) G A holds for each
B^B0.
37
Proof. If / : X —» Y is measurable with respect to A and B then f~x(B) G A
for each B € B0 because B0 C B. Now assume that G A holds for each
B e B0. Let T be the collection of all subsets B of Y such that f~1(B) <E A. Since
f f 1(B°) = (f~1(B))°, and / - 1 ( | J n B n ) = ( J w e have that
J~ is a cr-algebra on Y. Since J~ includes B0 it must include B. Thus f is measurable
with respect to A and B. •
Proposition 2.5.5. Let X and Y be Hausdorff topological spaces, and let f : X —>
Y be continuous. Then f is Borel measurable.
Proof. Since f is continuous, f 1 (V ) is an open subset of X for each open subset
U of Y. Since the collection of open subsets of Y generates B(Y^j, the measurability
of / follows from Proposition 2.5.4. •
Proposition 2.5.6. Let (X,A) be a measurable space, and let Y be metrizable
topological space. Then a function f : X ~^Y is measurable with respect to A and
B{Y) if and only if for each continuous function g : Y —y R the function g o f is
A-measurable.
Proof. If f is measurable with respect to A and B(Y), then the measurability of
g o / for each continuous function g follows from Proposition 2.5.5 and Proposition
2.5.3.
Now assume that for each continuous function g \ Y —y 1R the function g o f is
A-measurable, and let d be a metric that metrizes Y. Suppose that U is an open
subset of Y. There is a continuous function gu : Y —> R such that
U = {y e Y\gu(y) > 0}.
For instance if U / Y define gu by gu(y) = d(y, Uc), and if U = Y let gu be such
that gu{y) — 1. The set / - 1 (U) is equal to
{x e X : (gu o f)(x) > 0}
38
and therefore belongs to A because gu o f is ,4-measurable. Therefore / is measur-
able. •
Propos i t ion 2.5.7. Let (X,A) be a measurable space, letY be a metrizable topo-
logical space and for each positive integer n let fn : X —> Y be measurable with
respect to A and B(Y). If limn fn{%) exists for each x 6 X then the function
f '• X -» Y given by f(x) — limn fn(x) is measurable with respect to A and B(Y).
Proof. Note that if g : Y -»• R is continuous, then g(f(x)) = limn g(fn(x)) holds
for each x in X. For each n, g(fn(x)) is measurable by Proposition 2.5.6. Also
since each g(fn{x)) is real valued and measurable, lim„ g(fn{
x)) is also measurable.
Since this implies g(f(x)) is measurable, again applying Proposition 2.5.6 gives us
that / is measurable. •
If (X, A) is a measure space and C is a subset of X, then
Ac = {A fl C where A £ A}
is a a-algebra of subsets of C.
Proposi t ion 2.5.8. Let (X,A) be a measurable space, let Y be a Polish space, and
for each positive integer n let fn : X —>• Y be measurable with respect to A and
B(Y). Let
C = {x G X\\imfn(x) exists}.
n
Then C £ A. Furthermore, the map f : C —$• Y defined by f ( x ) = limra fn{x) is
measurable with respect to Ac and B(Y).
Proof Let J be a complete metric for Y. Then C is the set of those x G X for
which (fn(%))'^L1 is a Cauchy sequence in Y. We need to show that for each positive
integer n the set
1 1 An = [(yi,y2)€Y xY d(yi,y2) < n
39
is open in Y xY. Let n G N and let (x,y) G An. Choose e > 0 such that
d{x, y) + e < Let (p, 5) 6 f e) x Bd{y, |e) . Then
9) < <*(p,«) + d(x, y) + d(y, 2)
1 u ^ 1
< -e + d(x,y) + - e
1 < n
Consequently (x,y) G \e) x Bd(y,\e) C and is open. Therefore
An G H(F) x B(y) by Proposition 2.5.1. For each i,j,n define the set C(i,j, n) by
C(i,3,n) = {* e X d(fi(x), f j ( x ) ) < i } .
Define the function faj : X ->• Y x Y such that
(jM^) = CftO), /?(»)•
Suppose A and B are open in Y. Then
4>~HAxB) = { - ' ( A ) n f j \ B ) .
Since both / j and f j are measurable <f>ij is also measurable. We now show that
C{i,j,n) = An). First let x G C(i,j,n); then
Vi(x),fj(x)) < -•
n
Therefore (/,- {x), f j ( x ) ) G An. This implies that z G <t>J(An) and C(i,j,n) C
<i>ij(An)• Now let z G ^{An). Then <j>ij{z) = ( f i ( z ) J j ( z ) ) G An. This implies
that d(fi(z),fj(z)) < ± and 2 G C(i,j,n). Therefore C(i,j,n) = (f)^1 (An). Since
4>ij is measurable and An is open C(i,j,n) G A.
The fact that C(i,j, n) G A will be used to show that C G A. First we need to
show that c = n u n n n k i>kj>k
40
First let x G C. From above we have that ( / n (^) )^L 1 is a Cauchy sequence in Y. Let
n G N. There exists a positive integer k such that if z, j > k then d(fi(x), f j ( x ) ) <
Since this is true for all such n
Now suppose
* 6 nu n n c(^»-n k i>kj>k
x g nu n n c^^n)-n k i>kj>k
For each n € N there exists a k such that if i,j > k then d(fi(x), f j ( x ) ) <
Therefore ( / n (x))^L 1 is a Cauchy sequence. This implies that x € C and
*=nunn<™,™). n k i>kj>k
Consequently since each C(i,j,n) G A and A is a cr-algebra, C e A.
Let / : C —> Y be defined by f ( x ) — limn fn(x). Then by Proposition 2.5.7 we
have that f ( x ) is measurable with respect to Ac and B(Y). •
Since Ac is a subset of A it is also true that / is measurable with respect to A
and B{Y). A similar argument to that above is made in the following proof.
Propos i t ion 2.5.9. If (X, A) is a metrizable space, Y is a separable metrizable
space, and f,g\X-^Y are measurable with respect to A and B(Y), then
{x G X\f(x) = g(x)}
belongs to A.
Proof. Let d be a metric for Y. For each positive integer n let
d(yi,y2) < -n
= <(yi,y2) eY xY
41
The set An is open in Y x F . By Proposition 2.5.1, An G B(Y) x B(Y). Define the
set
1 1 d{f(x),g(x)) <
n Cn = G X
Define the function 4> : X —$Y x Y by
<A(®) = (f(x),g(x))-
Suppose A and B are open in Y; then,
r 1 ( i x B ) = r ' ( X ) n 9 - ' m .
Since / and g are measurable, <f> is also measurable. We assert that
Cn = 4>~l d(yuy2) < An).
Let x G Cn. Then d(f(x),g(x)) < i . Therefore ( / (a) , </(x)) G A n , and x G
4* l{An)- Now let a; G 4> 1(An). T h e n <f>(x) = (f(x),g(x)) G An; consequent ly ,
d(f ( x ) jg(x) ) < ^ and x G C*n. Therefore we have C n = <^ -1(An) as desired.
Since is in B(Y) x # ( F ) and 4> is measurable, Cn G A. Let C = Cn. If
x G C then d(f(x),g(x)) < i for all n, and / ( s ) = g(x). Therefore
c = {x € X\f(x) = g{x)}.
Since C = P|n C n , a countable intersection of elements from A, we have C G A. •
A positive measure on A is regular if
i) each compact subset K of X satisfies
fi(K) < oo,
ii) each set A in A satisfies
= inf{^/(Z7)|A C U and U is open},
42
iii) each open subset U of X satisfies
n(U) — sup{/u(i!i")|K C U and K is compact}.
We would, like to show that each finite Borel measure on a Polish space is regular.
To do this we need the following lemmas.
L e m m a 2.5.10. Let be a measure space. If (Ak)k is an increasing se-
quence of sets that belong to A, then fi({Jk Ak) — lim^ ji(Ak).
Proof. Define a sequence {Bn)n of sets by letting B\ = A\ and Bj = Aj — Aj-1 if
j > 1- Then the sets are disjoint and measurable. Note that ( J n A n =
U„ Bn and therefore
MlM-) = E ' W n n
n = lim y jjiBi
n < ^ n i—1
lim/u(l J Bi) n '-s
i= 1 ]im/j,(An)
as desired. •
L e m m a 2.5 .11. Let X be a Hausdorff space in which each open set is an J-a, and
let fj, be a finite Borel measure on X. Then each Borel subset A of X satisfies:
(1) fJ^(A) = i n f : A C U and U is open},
and
(2) fi(A) = sup{^t(.F) : F C A and F is closed}.
Proof. Let C be the collection of all subsets A of X that satisfy (1) and (2). We
begin by showing that C contains all open subsets of X. Let V be open in X. Then
obviously V satisfies
/^(V) — inf{/n(U) : V C U and U is open}.
43
By hypothesis V is an FA. Let (CN)N be a sequence of closed subsets of X such that
V = (Jn CN. We can assume without loss of generality that (CN)N is an increasing
sequence, because if necessary we could replace each CN with (Jj=i CJ • Then by
Lemma 2.5.10, /J,(V) = limny«(Cn). It follows that
pt-iy) = sup{FJ-(F) : F C y and F is closed}.
Consequently C contains all the open subsets of X. Now let A be a, Borel set
such that for each e > 0 there exists an open set U and a closed set C such that
C C A C U and /J(U — C) < e. Since A — C C U — C and U — A C U — C we have
FJ-(A — C) < N(U — C) < e and (x{XJ — A) < /J,(U — C) < e. Therefore
n{A) < fi(A n C) + v{A n CC) = n(C) + v{A -C)< /J(C) + e,
and
fi(U) < n{U n A) + fi(U n Ac) = n(A) + - A) < fi(A) + e.
Consequently A satisfies (1) and (2) and therefore is in C. Now we can show that
C is a er-algebra. Clearly C contains X because X is open. If A 6 C and e > 0
there exist C and U which are respectively closed and open such that C C A C U
and FJ,{U — C) < e. Then UC and CC are respectively closed and open and satisfy
U° C Ac C Cc and n(Cc — U°) < e. Therefore it follows from above that Ac £ C.
Now let (Ak)k be a sequence of sets in C and let e be a positive number. For each
k choose a closed set Ck and an open set Uk such that Ck C Ak C Uk and
TIVU - CK) <
Let U = \JKUK and C = |Jfc CK. Then U and C satisfy the relations C <Z\JKAK CU
and
fi{u ~C)< ,x(U(Uk - Ck)) < - ck) < 6.
44
The set U is open but the set C can fail to be closed. However for each n the
set U L i is closed and ((Jfc=i Ck)n is an increasing sequence. Therefore using a
similar argument to that in Lemma 2.5.10 we have
n
v(U-C) = limv(U- (J Ck). n
k=1
Consequently we can find an N G N such that fJ.{U — UfcLi @k) < e- This gives that
UfcLi Ak G C, and it follows that C is a <r-algebra. Since C contains all the open
sets, B(X) C C. •
P r o p o s i t i o n 2 .5 .12 . Every finite Borel measure on a Polish space is regular.
Proof. Let X be a Polish space, let d be a complete metric on X, and let /i be a
finite Borel measure on X. Since X is metrizable, X is Hausdorff. Let U be an
open set in X. By Proposition 2.4.1 the set X — U is a Gs. Therefore U is a
By Lemma 2.5.11, each Borel set A satisfies
/i(A) = i n f { n ( U ) : A C U and U is open},
and
jJ.{A) = sup{|«(i?1) : F C A and F is closed}.
Therefore we only have left to show that
yu(̂ 4) = sup{/i(liQ : K C A and K is compact}.
First consider the case where A — X. Let (xk)k be a sequence whose terms form a
dense subset of X. Let e > 0. Note that for each n, X = U ^ ! Bd(xk, J ) . For each
n G N choose kn such that
k
M U B J ( x l , h ) > / J ( x ) - ^ . k=1
45
Define the set K by oo k IVn
« = n u z > -n=1fc=l U
The set K is closed and because d is complete in X, K is also complete. Now we
need that K is totally bounded. Let e > 0 and choose n such that ^ < | e . Clearly
{Bd{xki \)}\=i is a finite covering of K. For each fc, Bd{xk, ^ Bd(xk, je) . Let
y be a limit point of Bd{xk^~). Choose S > 0 such that ^ + S < | e . There exists
z G Bd{xk, ^ ) such that z G Bd(y,S). Then
d(xk,y) < d(xk,z) + d(y,z) < - + S < ]-e. H/ ZJ
Therefore Bd(xk, C Bd(xk, | e ) . Now we have that {Bd(xk, is an open
covering of K. For each k = 1,2, ...kn choose, if possible, yk G K such that yk G
Bd(xk, | e ) . Let
A = {k : y}~ was chosen}.
Clearly A is finite. Consider {Bd(yk,e)}kea- Let z £ K and fix k G {1,2,. . . , kn}
such that z G Bd(xk, | e ) . Since z £ K and z G Bd(xk, | e ) it was possible to find
yk G K such that yk G Bd{xk,\^)- Consequently, k G A. Therefore
d(Vk,z) < d(yk,xk) + d(xklz) < e.
Consequently z G Bd(yk,e)- Now we have that {Bd(yk,e)}k£a is a finite covering
of ET. Since each yk G K for all k G A, we have that K is totally bounded. Now we
have that K is compact. Furthermore,
oo kn ^ oo
M X - K ) < - ((J -))) < £ ^ = e, n= 1 k=1 rc=l
and therefore n{K) > / i (X) — e. Since e is arbitrary we have
fJ.{X) = sup{yu(isf) : K C X a n d K is compact}.
46
Now let A be an arbitrary Borel subset of X and let e be a positive number.
Choose a compact set K such that (J,(X — K) < t. We have that
fj-(A) = sup{/-t(F) : F C A and F is closed}.
Choose a closed subset F of A such that fJ,(F) > /J,(A) — e. Then K C\F is a compact
subset of A and
H {K DF)> n{F) - n(F n {X - K))
> fi(F) - n{X - K)
> — e — e
> jji(A) — 2e.
Since e is arbitrary
^{A) = sup{^(K) : K C A and K is compact}.
Thus fi is regular. •
CHAPTER III
ANALYTIC SETS
3.1 Borel Subsets of Polish Spaces
Now we turn our attention to analytic sets. Recall from the introduction that
a subset A of X is analytic if there is a Polish space Z and a continuous function
/ : Z —> X such that f ( Z ) = A. The purpose of this section is to show that every
Borel subset of a Polish space is analytic. Later we will see that there are analytic
sets which are not Borel.
Propos i t ion 3.1.1. Let X be a Polish space. Then each open subset and each
closed subset of X is analytic.
Proof. By Proposition 2.1.4 each closed subset and each open subset of X is Polish.
Let Y be a closed (or open) subset of X. Define / : Y —> X such that f ( y ) = y
for all y E Y. Let U be an open set in X. Then y E f~l{U) if y E U and y E F .
Also if z E U fl Y, f ( z ) = z E U] therefore / - 1 (?7) = U Pi Y which is open in Y.
Therefore / is continuous. Clearly f ( Y ) = Y. Consequently Y is analytic. •
Propos i t ion 3.1.2. Let X be a Polish space, and let A i ,A2 , ... be analytic subsets
of X. Then (Jft Ak and Ak are analytic.
Proof. For each k choose a Polish space Zk and a continuous function fk '• Z^ X
such that fk(Zk) = Ak- Let Z be the disjoint union of the spaces Z\, Z2, Z3,... and
define / : Z —»• X so that for each k it agrees with fk on Zk- Then Z is Polish by
Proposition 2.2.4. Let U be an open set in X. Then
r\u) = f-\u n AO u r\u n A2) u ... u f-\u n An)...
48
= / f 1 {u n Aj) u f ; \ u n A2) U ... u / " ' ( £ / n An)...
= / f 1 ( C f ) u / 2 - , ( t / ) u . . . u
Therefore / _ 1 (C / ) is open and / is continuous. Let y G f{Z). Then there exists a
positive integer k such that y G f(Zk) — A^. Therefore y G (Jk-^-k- Now suppose
z € Ufc Ak- Then z G Ak for some k, giving 2; G f(Zk) and z G f(Z). Therefore we
have that f ( Z ) = (Jfc Ak and \Jk Ak is analytic.
Next form the product space Zk and let A consists of those elements (zk)k i n
Zk for which fi(zi) = f j ( z j ) holds for all i and j. Let (pk)k be a limit point of A.
Let (yi)i be a sequence from Zk which converges to (pk)k, where yi = (yii, 2/i2, - --)
for each i. Then (yij)f^i converges to p j for all j. Since f j ( y i j ) = fk{lfik) f ° r all j and
k, we have f j ( p j ) = fk{Pk) f ° r all k. Thus (pk)k € A and A is closed. By Proposition
2.3.3, Zk is Polish and by Proposition 2.1.4, A is Polish. Define g : A —> X such
that g((zk)k) = /1 (z\ )• Then the continuity of f\ implies that g is also continuous.
Let x G g(A). Then there exists (yk)k G A such that g((yk)k) = fi{yi) — x-
Therefore x G Uk for all k because fk{yk) — f\{y\) = x and fk{Zk) = Uk for all k.
Consequently g(A) C C\kUk- Now suppose y G HkUk- Then for each k there exists
Zk G Zk such that fk(zk) — %• Consequently (Zk)k G A and g((zk)k) = x. Now we
have that (/(A) = ClkUk and that f\kUk is analytic. •
L e m m a 3.1.3. Let X be a Hausdorff topological space. Then B(X) is the smallest
family of subsets of X that
(a) contains the open and the closed subsets of X,
(b) is closed under the formation of countable intersections,
(c) is closed under the formation of countable disjoint unions.
Proof. Let S be the smallest collection of subsets of X that satisfy (a), (b), and
(c). Note that we can find S by taking the intersection of all such collections. Let
49
S0 = € S and AC £ <S}. It is clear that S0 C S C B(X). Thus we need to
show that S0 is a er-algebra that contains each open subset of X and it will follow
that S0 = = B{X).
It is obvious that S0 contains the open subsets of X and that SQ is closed under
complementation. Now suppose that (AN)N is a sequence of sets in S0. Then (Jn AN
is the union of the sets
AI,A\ n AI,A\ n AC2 n A$, ...
Since these sets are disjoint and belong to S, (JN AN must also belong to S. Further-
m o r e (Un AN)° i s the intersection of a sequence of sets in S, and must belong to <5.
Consequently | J n AN belongs to S0. It follows that S0 is closed under the formation
of countable disjoint unions. Therefore S0 is a cr-algebra, and S0 = S = B(X). •
Proposit ion 3.1.4. If X is a Polish space, then each Borel subset of X is analytic.
Proof. By Propositions 3.1.1 and 3.1.2 the class of analytic subsets of X satisfy
conditions (a), (b), and (c) of Lemma 3.1.3. Therefore B(X) is a subset of the
analytic sets and each Borel subset of X is analytic. •
3.2 The Space Af
Recall from an earlier example that M is the space Nn . This space is very
important in the study of analytic sets. In fact, we will show in this section that
each analytic set is the image of Af under some continuous function. This section
also provides many unique methods of producing the continuous functions needed
to show that a set is analytic. First we need to determine some more properties of
analytic sets, and we will also find that each Polish space is the image of M under
some continuous function.
50
Proposit ion 3.2*1. Let Xi,X2,... be a finite or infinite sequence of Polish spaces;
and for each k let Ak be an analytic subset of Xk- Then ]Jk A* is an analytic subset
ofUk^k.
Proof, For each k choose a Polish space Zk and a continuous function fk : Zk —> Xk
such that fk(Zk) — Ak. Define a function / : zk -)• n k X k by f{(zk)k) =
(fk(zk))k- We have that Zk is Polish by Proposition 2.3.3, and / is continuous
by Proposition 2.3.1. Clearly Zk) = Ilfc Ak. Therefore Ak is analytic. •
Lemma 3.2.2. Let X be a Hausdorff topological space, and let Y be a subspace of
X. Then
B(Y) — {A\ there is a set B in B(X) such that A = B f ) Y } .
Proof Let B{X)y denote the collection of subsets of Y that have the form B D Y
for some B in B(X). We need to show that B(Y) = B(X)Y. Let / : Y -> X be
a function such that f ( y ) = y for each y 6 Y. Then / is continuous and hence
measurable with respect to B{Y) and B(X). Since each subset B of X satisfies
/ 1(-®) —BOY, we have that B(X)y C B(Y). Now let us show that B(X)y is
a a-algebra containing the open subsets of Y. If U is open in Y by definition of
subspace topology U = V (1 X for some V open in X. Therefore B{X)y contains
all the open sets in Y. Suppose A € B(X)y. Then A = B fl Y for some B £ B(X).
Also
Y - A = Y ~{Bf\Y)
= Y n (BDY)C
= Yr\(BcU Yc)
= F n Bc.
51
Since B E B(X), B° E B(X). Hence Y — A E B(X)y• Now suppose is a
sequence of elements from B(X)y. For each i choose Bi such that At = BtC\Y and
Bi EB{X). Then
U = U(5i n r ) = ( | J 5 0 n r .
i i i
Since (J iBi E B(X) we have (J^ Ai E B(X)y. Therefore B(X)y is a cr-algebra
containing all open subsets of Y. This implies that B(Y) C B(X)y. We now have
shown that B(Y) = B{X)y. •
Proposition 3.2.3. Let X and Y be Polish spaces, let A be an analytic subset
of X, and let f : A —>• Y be Borel measurable (that is measurable with respect to
B{A) and B(Y)). If A\ and A2 are analytic subsets of X and Y respectively, then
f ( A f ) A i ) and f~1(A2) are analytic subsets ofY and X respectively.
Proof. Suppose the hypotheses are satisfied. Proposition 2.5.2 implies that gr(/) E
B(A x Y) and Lemma 3.2.2 then implies that there is a Borel subset B of X xY such
that gr( / ) = B D {A x Y"). We have that Y is analytic by Proposition 3.1.4. Also
gr( / )n(A 1 x F ) is analytic by Proposition 3.1.2 and 3.2.1. Similarly, g r ( / )n(A 1 x Y)
is an analytic subset o f X x F and therefore is the image of a Polish space Z under
a continuous map, say h. Let Try be the projection o f l x F onto Y. Then tt y is
continuous and 7Ty o h is a continuous map from Z into Y.
We want to show that (7xy o h)(Z) = f(A Pi A\). We will show this by inclusion
in both directions. Let z E {TCy o h)(Z). Then there exists an x E X such that
{x,z) E h(Z) = g r ( / ) D x Y). Then x E Ax and f{x) = z E Y. Since
the domain of / is A we have that x E A. Consequently z E f(A f| Ai) and
(Try o h)(Z) C f(A n Ax). Now suppose y E f{A fl Ax). Then there exists an
x E AD Ax such that f(x) = y. Then (x,y) E gr ( / ) fl (Ax x Y). This implies
that there exists z E A such that h(z) = (x,y) and (Try o h)(z) = y. Therefore
52
f(A n i l ) C (7Ty o h)(Z). Now we have that f(A fl A\) = (7ry 0 h)(Z), ixy o h is
continuous, and Z is a Polish space. Consequently / ( A fl Aj) is analytic.
The argument is similar to the above that gr( /) fl (X X A2) is also an analytic
subset of X xY. Now let h be a continuous function and Z be a Polish space such
that h maps Z onto gr( /) fl (X x A2). We need to show that / - 1 {A2) = ( f f x oh)(Z).
Let x G f~1(A2). Then x G A and f(x) = y for some y G A2. This implies that
(x>y) € gr ( / ) fl (X x A2). Then there exists z G Z such that h(z) = (x,y), which
implies (nxoh){z) = x. Therefore C ( t txo/i)(Z). Now let x G (nx°h)(Z).
Then there exists y G Y such that (x,y) G h(Z). This implies that f(x) = y and
y G A2. Consequently x G f~1(A2) and / _ 1 ( A 2 ) = (nx o h)(Z). Since ( irx o h) is
continuous and Z is Polish, we have that f~1(A2) is analytic. •
The following proposition uses a method of constructing a continuous function
that will be useful again later in this study. The function is constructed by first
constructing a net of nested closed sets and applying to each chain the Cantor
Nested Set theorem to find the unique point in the intersection of sets in the chain.
The continuous function sends a sequence used to index the closed sets to that
unique point. In the following proof, if A is a subset of the topological space X
denote the interior of A by A0.
Proposit ion 3.2.4. Each non-empty Polish space is the image of Af under a con-
tinuous function.
Proof. Let X be a non-empty Polish space and let d be a complete metric for X.
We shall construct a family {C{nx, n2 , . . . , nk)} of subsets of X, indexed by the set
of all finite sequences (n\, n2,..., n^) of positive integers. We do this by induction
on k. First suppose k = 1, and let be a sequence whose terms form a
53
dense subset of X. Then for each n j we can define
C(n\) = {x G X\ d(x,xni) < 1}.
Now suppose k = 2 and let (£n2)n^=i be a sequence whose terms form a dense
subset of Let
C (n i , n 2 ) = | x G C(ni) d(x,xn2) < ^
Then C(ni) — [Jn2 C(n\, 712). If we continue this process inductively, then the
following conditions hold:
(a) C(ni , n 2 , r i f c ) is closed and non-empty,
(b) the diameter of C(n\,n2, ...njt) is at most £,
(c) C(ni , n2 , . . . , nfc-i) = (Jnfc C(n 1 ; n2 , ...njt), and
(d) ^ = U „ 1 ^ ( » i ) .
We turn to the construction of a continuous function that maps J\f onto X.
Let n = (n/c)'^L1 be an element of Af. It follows from (a), (b), and (c) above
that C(n1),C(n1,n2),C(n1 , n2, n3),... is a decreasing sequence of non-empty closed
subsets of X whose diameters approach zero. Thus there is a unique element in the
intersection of these sets by Theorem 2.4.2 and we can define a function / : Af —> X
by letting / ( n ) be the unique member of p|fc C(n j , ..., n^). Note that if m and n
are elements of Af such that m; = m holds for i = 1,2,..., k then d(f( m), /(n)) < \.
It follows that / is continuous. Finally (c) and (d) above imply that for each x in X
there is an element n e Af such that x G D* C(nun2,...,nk) and hence x = /(n);
thus / is surjective. •
Corollary 3.2.5. Every non-empty analytic subset of a Polish space is the image
of Af under some continuous function.
54
Proof. If A is the image of the Polish space Z under the continuous function / , and
Z is the image of Af under the continuous function g by Proposition 3.2.4, then A
is the image of Af under the continuous function f o g . •
The following is a similar result that will be useful in a later section.
Propos i t ion 3.2.6. Let X be a Polish space. A subset A of X is analytic if and
only if there is a closed subset of Af X X whose projection on X is A.
Proof, Since Af is Polish, Af x X is Polish by Proposition 2.3.3. Any closed subset
of Af x X is Polish by Proposition 2.1.4. If A is the projection of a closed subset of
Af x X onto X , then A is analytic because the projection function is continuous.
Now suppose that A is an analytic subset of X . If A is empty, then it is the
projection of the empty subset of Af x X onto X. Otherwise by Corollary 3.2.5
there is a continuous function / : Af -> X such that f ( A f ) = A. Recall that
g r ( / ) = {(Xjy) e Af x X\y = f(x)}
is closed. Since
7 r x(gr( / ) ) = {y\y = f ( x ) for some x £ Af}
we have 7rx(gr(/)) = A. •
3.3 Zero Dimensional Spaces
A topological space is zero-dimensional if its topology has a basis that consists
of sets that are both open and closed. Among the zero-dimensional spaces are the
space of all rational numbers, the space of all irrational numbers, and each space that
has the discrete topology. Note that a product of zero-dimensional spaces is zero-
dimensional, that a subspace of a zero-dimensional space is zero-dimensional, and
that the disjoint union of a collection of zero-dimensional spaces is zero-dimensional.
55
In particular, the spaces J\f and {0,1}N are products of zero-dimensional spaces and
so are zero-dimensional. We need to show that each Borel subset of a Polish space is
the image under a continuous injective map of some zero-dimensional Polish space.
To do this we need the following lemma.
L e m m a 3 .3 .1 . Every separable metrizable space is homeomorphic to a subspace of
[0,1]N, and every Polish space is homeomorphic to a Gs in [0,1]N.
Proof. Let d be a metric for the separable metrizable space X. Let (xn)n be a
sequence whose terms form a dense subset of X. Consider the map / from X to
[0,1]N that takes the point x to the sequence whose nth term is min{l , d(x, xn)}.
Let A be the image of X under / . Clearly f : X —>• A is onto. Let y and z be
distinct points in X. Let
6 = min j £ < % * ) , 1 J .
Choose n £ N such that xn E Then d(y,xn) < 8. However,
d(z,xn) + d(xn,y) > d(y,z),
and
d(z, xn) > d(y, z) - d(xn,y) >2S- d(xn,y) >26 - S = 6.
This implies that the coordinate of / ( y ) is not equal to the n c o o r d i n a t e of
f ( z ) and f ( y ) ^ f ( z ) . Now we have that / is one-to-one.
Let (zi)i be a sequence in X which converges to y. Clearly f ( y ) e A. Let n G N
and e > 0. There exists N e N such that d(Zi,y) < e for all i > N. Suppose i > N.
Then
d(yZi, xn) ^ d(y, xnJ -(-
and
56
d(zi,xn) - d(y,xn) < d(y,Zi).
Similarly,
d(y, xn) < d(zi,xn) + d(y, Z{),
and
d(y,xn) - d(zi,xn) < d(y,Zi).
This implies that \d(y,xn) — d{zi,xn)\ < d(zi,y) < e. Therefore the nth coordinate
of ( f ( z i ) ) i converges to the nth coordinate of / (y) . Since n was arbitrary, / is
continuous.
Now suppose (f(zi)) converges to f ( y ) in A. Let e > 0, and choose n such that
d(y,xn) < e. Choose N such that for all i > N we have
\d{xn,Zi) - d(xn,y)| < e.
Then
d(xn,Zi) - d(xn, y) < e,
and
d(xn, Zi) < e + d(xn,y).
This implies that
d{zi, y) < d(xn, z^ + d(xn, y) < e + 2 d ( x n , y ) < 3e.
Therefore (zj)j converges to y, and / is a homeomorphism. In addition, if we
suppose that X is a Polish space the fact that / is a homeomorphism along with
Proposition 2.4.4 implies that A is a Polish space. Since A is Polish it must also be
a Gs in [0,1]N. •
57
Proposit ion 3.3.2. Each Borel subset of a Polish space is the image under a
continuous injective map of some zero-dimensional Polish space.
Proof. We begin by showing that each Polish space is the image under a continuous
injective map of some zero-dimensional Polish space. First consider the interval
[0,1]. It is the image of the space {0,1}N under the map F : {0,1}N —> [0,1] that
takes the sequence (xk)k to the number | £ . Each number in [0,1) that has two
binary expansions (that is, each number in (0,1) that is of the form ^ for some
m and n) is the image under F of two elements of {0,1 }N: the remaining members
of [0,1] are images of only one element of {0,1}N. Thus if we remove a suitable
countably infinite subset from {0,1}N, the remaining points form a space Z such
that the restriction of F to Z is a bijection of Z onto [0,1]. A similar argument to
that which showed that the map from {0,1}N into the Cantor set is continuous can
be made to show that F is continuous. Since {0,1}N is zero-dimensional, we have Z
is zero-dimensional. Since {0,1}N — Z is countable and consequently an we have
that Z is a G& and therefore Polish. Hence [0,1] is the image under a continuous
injective map of a zero-dimensional Polish space. It follows that [0,1]N is the image
of the zero-dimensional Polish space under a continuous injective map.
Now suppose that X is an arbitrary Polish space. By Lemma 3.3.1 there is a
homeomorphism G of X onto a subspace of [0,1]N such that G(X) is Polish and a
Gs• Let H be a continuous injective map from ZN onto [0,1]N. Then H^1(G(X)) is
a Gs in and hence is also Polish. Let H0 be the restriction of H to H~1(G(X)).
Then X is the image of the zero-dimensional Polish space H~l(G(X)) under the
continuous map G ' o H0.
We turn to the Borel subsets of X. Let F consists of those Borel subsets B of X
for which there is a zero-dimensional Polish space Y and a continuous injective map
f '• Y —$• X such that f(Y) = B. The first part of this proof along with Proposition
58
2.1.4 implies that T contains all open and closed subsets. A modification of the
proof of Proposition 3.1.2 shows that T is closed under the formation of countable
intersections and countable disjoint unions. Thus Lemma 3.1.3 implies that T —
B{X). •
Propos i t ion 3.3.3. Let X be a zero-dimensional separable metric space, let U be
an open and non-compact subset of X, and let e be a positive number. Then U is
the union of a countably infinite family of disjoint sets, each of which is non-empty,
open, closed, and of diameter at most e.
Proof. Since U is open and non-compact there is a family A that consists of open
sets and whose union is U, but A has no finite subfamily whose union is U. Let
V be the collection of all subsets of X that are both open, closed, of diameter at
most e, and included in some member of A. Since X is zero-dimensional the set U
is the union of the family V. Since X is second countable there exists a countable
subfamily of V, call it V0, whose union is U. List the sets in V0 in the sequence
Vi, V2,... and consider the non-empty sets that appear in the sequence
vu v{ n v2, vf n v2c n y3,...
It is clear that these sets are open, closed, disjoint and of diameter at most e, and
that their union is U. If all but a finite number are empty a finite sub collection of
A would cover U. Therefore there must be a countably infinite number. •
3.4 Non-Borel Analytic Sets
In this section we will show that there does exists a non-Borel analytic set. To do
this we first need to study some properties of uncountable Borel subsets of Polish
spaces. A point x of X is a condensation point of A if every open neighborhood of x
59
contains uncountably many points of A. The following lemma about condensation
points will be useful in our study of uncountable Borel subsets.
Lemma 3.4.1. Let X be a separable metrizable space, and let C be the set of
condensation points in X. Then C is closed and Cc is countable.
Proof. Let W be a countable basis for X. Then x (Jz C if and only if there is a
countable open set that belongs to U and contains x. Hence C° is the countable
union of countable open sets and is therefore open and countable. This implies that
C is closed. •
Proposition 3.4.2. Let X be a Polish space, and let B be an uncountable Borel
subset of X. Then there is a continuous injective map f : Af —> X such that
/O^O B and such that B — f(M) is countable.
Proof. There is, according to Proposition 3.3.2, a zero-dimensional Polish space Z
and a continuous injective map g : Z —» X such that g(Z) = B. Thus it will suffice
to construct a continuous injective map h : J\f Z such that Z — h(Af) is countable,
and then to define / to be g o h.
Let Z0 be the collection of all points in Z that are condensation points of Z. Then
by Lemma 3.4.1 and by Proposition 2.1.4, ZQ is zero-dimensional. Also, by Lemma
3.4.1, Zr0 is countable. If x 6 Z0 every neighborhood of x contains uncountably many
points of Z. Since Z'0 is countable every neighborhood of x contains uncountably
many points of Z0. Therefore every point of Z0 is a condensation point of Z0.
Suppose that d is a complete metric on Z0. For each k we construct a family of
sets indexed by as follows. Choose a0 £ Z0 and let U = ZQ — {a0}. Since aQ
is a condensation point of ZQ and therefore a limit point of Z0 we have that U is
not closed and consequently is not compact. Let e = 1. Then by Proposition 3.3.3
we can find a sequence of sets ( A ( n i ) ) ^ = 1 which are disjoint, non-empty, open,
60
closed, and of diameter at most 1, such that | J ^ _ 1 A(ni) = U. Also each A{rt\)
consists entirely of condensation points of itself. We can repeat this construction
producing for each k and n\, n2, sets A{n\, n 2 , n - f c ) , rik — 1,2,..., that are
disjoint, non-empty, open, closed and of diameter at most and are such that
U ^ = i ^(n ~~ ^ -~nk) A(rii, ...,rik-i) less some point a(ni , . . . , n ^ - i ) . Since each
point in A{n\, n 2 , •••> i s a limit point and
a(ni,.. . ,njfe_i) £ |J A(n1,n2, ...nk), nk = 1
the set U ~ = i ^ ( n i ? n 2 ) is n o t closed and therefore not compact.
Define h : M —> Z by letting h(n) be the unique point in D^. 1A(ni ,n2, . . . ,nfe) .
Note that for each m and n in M such that m; = rti hold for i = 1,2,.. .k , we have
d(h{m)i h(n)) < j. It follows that h is continuous. Also, since A(ni,n2, . . . ,n.fc-i ,p)
and A(ni, n 2 , . . . , n j t - i , <?) are disjoint if p ± q it follows that h is injective. Suppose
that x is not a point removed in the above construction. Then x G A(n\) for
some n\. Since x ^ a (n i ) we have x G A(ni ,n2) for some n-2- Inductively for
each k there exist some choices n\, 712,such that x G A(ni , n 2 , . . . , ^fe). This
implies that there exists an n G J\f such that h(n) = x. Consequently Z0 — h(N)
is the countably infinite set consisting of points removed during the construction
of the sets A(ni,ri2, ...,njt). Since Z% is countable it follows that Z — h(Af) is also
countable. Since g(Z) = B, if we define f = g o h, f(Af) C B. Suppose 2; is in the
set B — f{M). Then z G g(Z) and 2: ^ g(h(Af)). Consequently there exists x G Z
such that g(x) = z. Then x G Z — h(Af). Since g is injective and Z — h(M) is
countable, B — f(Af) must also be countable. •
Corollary 3.4.3. Each uncountable Borel subset of a Polish space includes a subset
that is homeomorphic to {0,1}N.
Proof. Let X be a Polish space, and let A be an uncountable Borel subset of X.
61
Proposition 3.4.2 provides a continuous injective map / : M —• X such that / ( f i f ) C
A. Regard {0,1}N as a subspace of M and restrict / to {0,1}N. Then since {0,1}N
is compact this restriction is a homeomorphism of {0,1}N onto the subset / ({0,1}N)
of A. •
Since we found earlier that {0,1}' ' is homeomorphic to the Cantor set, the above
corollary also implies that each uncountable Borel subset of a Polish space includes a
subset that is homeomorphic to the Cantor set. Let us introduce some terminology
and notation. Suppose that X and Y are sets and that E is a subset of X x V.
Then for each i £ l and each y € F the sections Ex and Ey are the subsets of Y
and X given by
Ex = {yeY : (x,y) e E},
and
Ey = {x e X : {x,y) € E}.
Now let X be a set and let ZF be a family of subsets of X. A subset A of J\f x X is
universal for T if the collection of sections {j4n |n € N}, is equal to F.
Proposi t ion 3.4.4. Let A be an uncountable analytic subset of the Polish space X.
Then A has a subset that is homeomorphic to {0,1}N. Also, A has the cardinality
of the continuum.
Proof. Let A be an uncountable analytic subset of the Polish space X and let d
be a complete metric for X. By Corollary 3.2.5 there exists a continuous function
/ : M -> X such that f ( N ) = A. Choose a subset S of M such that the restriction
of / to S is a bijection of S onto A. We can do this by choosing for each a G A
exactly one n 6 Af such that / ( n ) = a. Let S0 consist of the points in S that are
condensation points of S. By Lemma 3.4.1 the set S — S0 is countable. Therefore
S0 is uncountable and we can choose «(0) and ®(1) from SQ such that x(O) ^ x(l).
62
Let 5 ( 0 ) = m i n { 1, |c?(a;(0), x ( l ) ) } . L e t
5 ( 0 ) = 5 ^ ( 0 ) , 5 ( 0 ) ) ,
a n d
5(1) = Bd(x(l), 5(0)).
T h e n 5 ( 0 ) n B{ 1) = 0. C h o o s e x ( 0 , 0 ) a n d x ( 0 , 1 ) in B{0) n S0 s u c h t h a t x ( 0 , 0 ) ±
x ( 0 , 1 ) a n d let
5 ( 0 , 0 ) = m i n | ^ ^ c ? ( a : ( 0 , 0 ) , a ; ( 0 , 1 ) ) | .
A l s o c h o o s e a?(l, 0 ) a n d x(\, 1) i n 5 ( 1 ) fl S0 s u c h t h a t a : ( l , 0 ) ^ x ( l , 1) a n d le t
5 ( 1 , 0 ) = m i n j i , ^ d ( z ( l , 0 ) , a ; ( l , 1 ) )
T h e n le t
B(0,0) = Bd(x(0,0), 5(0,0)) n £ ( 0 ) ,
B{0,1) = Bd(x{0,1), 5(0,0)) n £ ( 0 ) ,
5 ( 1 , 0 ) = Bd(x(l, 0 ) , 5 ( 1 , 0 ) ) n 5 ( 1 ) ,
a n d
5 ( 1 , 1 ) = 5 ^ ( 1 , 1 ) , 5 ( 1 , 0 p B ( l ) .
S i m i l a r l y for e a c h ( n i , r i 2 , r i k ) G { 0 , l } k c h o o s e x ( n i , « 2 , • ••«*;, 0 ) a n d
x(r i ] , t t .2 , • • • , « £ , 1) in So D B(ni,ri2, ...,rc.fc) s u c h t h a t
s ( n i , U2, ...nfc, 0 ) ^ x ( n i , n 2 , n * , 1)
a n d c h o o s e
5 ( n i , n 2 , . . . , n j t , 0 ) = m i n | ^ ^ ^, ^ < / ( « ( n i , n 2 , . . .n f c , 0 ) , x(ni, re2, n * , 1) j -
63
Let
B ( m , n 2 , ...,nfc,0) = Bd(x(n1,n2,...nk,0),S{n1,n2,...,nk,0)) n B(nun2, ...,nfc)
and let
B(nun2,...,nk,l) = , n2, ...nk, 1), 8(m, n2,..., n*, 0)) n B{nx, n 2 , "*)•
Define p : {0,1}N such that flf(n) is the unique element of n ^ = 1 B ( n 1 , n 2 ,
Note that if m and n are elements of {0,1}N such that mi = n j holds for i —
1,2,. . . ,k then % ( m ) , # ( n ) ) < It follows that g is continuous and / o g is
continuous. Again suppose that m, n € {0,1}N with m ^ n. Choose the smallest
positive integer i such that rn, / nt. Then without loss of generality we can
suppose that </(m) G jB(n i ,n 2 , . . . ,n i_ i , l ) and p(n) G B ( n i , n 2 , Since
_B(m,n2 , . . . ,n;_i ,0) n B(n i ,n 2 , . . . ,n , -_ i , l ) = 0 and / is injective on S we have
that ( / o <j)(m) 7̂ ( / o g){n) and that f o g is injective. Consequently {0,1}N is
homeomorphic to the image of {0,1}N under the map f o g .
Since / is a one-to-one function such that / { A f ) — A, we have that the cardinality
of A must be less than or equal to the cardinality of M. Since Af has the cardinality
of the continuum, the cardinality of A is less than that of the continuum.. By
the above, {0,1}N is homeomorphic to a subset of A. Since {0,1}N also has the
cardinality of the continuum, A must also have the cardinality of the continuum. •
Lemma 3.4.5. Let X be a separable metrizable space. Then there is an open subset
of M x X that is universal for the collection of open subsets of X, and there is a
closed subset of M x X that is universal for the collection of closed subsets of X.
Proof. Let V be a countable basis for X and let (Vn)n be a sequence whose terms
are the sets in V together with the empty set. Define a subset U of Af x X by
U = {(n,x) | x G Vnk for some k}.
64
Then U is open because it is the union of the open sets U(k,j), for some positive
integers k and j, defined by
U(k,j) = {n G Af\ nk = j} x Vj.
For each n in Af the section Un is given by
Un = {x G X\ (n ,x) G U} = k
Since each open set is the countable union of elements from V, we have that {?7n}n
is equal to the set of all open sets.
The complement of U is a closed subset of Af x X and is universal for the class
of closed subsets of X. •
Proposition 3.4.6. Let X be a Polish space. Then there is an analytic subset of
Af x X that is universal for the collection of analytic subsets of X.
Proof. Use Lemma 3.4.5, applied to the space Af x X, to choose a closed subset F
of Af x Af x X that is universal for the collection of closed subsets of Af x X. Let
A be the image of F under the map h that takes (m, n, x) (m, x). Then A is an
analytic subset of Af x X. Let m G Af. Then
Am = {x E X|(m, x) G A}.
Also,
Fm = {(n, x) G Af x X\ (m, n, x) G F}.
If (n, x) G Fm, then (m, n, a;) G F and (m, x) G A. This implies that x G Am.
Consequently Am is the projection on X of the corresponding section Fm of F.
Since F is universal for the closed subsets of Af x X, Proposition 3.2.6 implies that
the analytic subsets of X are exactly the projections on X of the sections Fm. Thus
A is universal for the class of analytic subsets of X. •
65
Corollary 3.4.7. There is an analytic subset of M that is not Borel.
Proof. According to Proposition 3.4.6, there is an analytic subset A of M x M that is
universal for the collection of analytic subsets of Af. Let S = {n G Af\ (n, n) G A}.
Recall that C = { (m,n) G Af X Af\ m = n} is a closed subset of Af x Af and is
consequently analytic. Also A fl C is analytic because it is the intersection of two
analytic sets. Since S is the projection of A fl C onto Af, S is also analytic. Now
suppose that S is a Borel set. Then S c is also a Borel set and is therefore analytic
by Proposition 3.1.4. Thus, since A is universal, there is an element n0 of M such
that Sc = A n o . Let us consider whether or not n0 belongs to Sc. If n0 G Sc then
the definition of S implies that (n0, n0) $ A, contradicting n0 G Sc = AUo. Likewise
if n0 ^ Sc, then (n0, n0) G A, contradicting n0 ^ Sc = Ano. In either case we have
a contradiction, therefore it must be that S is not a Borel set. •
CHAPTER IV
THE SEPARATION THEOREM AND ITS CONSEQUENCES
4.1 Borel Measurable Functions
Let X be a Polish space, and let and A2 be subsets of X. Then Ai and A2
can be separated by Borel sets if there are disjoint Borel subsets Bx and B2 of X
such that A\ C B\ and A 2 Q B2.
Theorem 4.1.1. Let X be a Polish space, and let Ai and A2 be disjoint analytic
subsets of X. Then A\ and A2 can be separated by Borel sets.
Proof. Let us begin by showing that:
a) if Ci, C 2 , a n d D are subsets of X such that for each n the sets Cn and D can
be separated by Borel sets, then U n C * a n d D c a n b e s e P a r a t e d B o r e l s e t s '
and
b) if E1,E2,..., and FUF2,... are subsets of X such that for each m and n the
sets Em and Fn can be separated by Borel sets, then jjm Em and [Jn Fn can be
separated by Borel sets.
First consider assertion (a). For each n choose disjoint Borel sets Gn and Hn
such that Cn C Gn and D C Hn. Then {JnGn and f]nHn are disjoint Borel sets
with Un °n Q Un Gn and DCf]nHn. Hence assertion (a) is proved.
Next consider assertion (b). Assertion (a) implies that for each m the sets Em
and \Jn Fn can be separated by Borel sets. Again applying assertion (a) we have
that U m Em and \Jn Fn can be separated by Borel sets.
We turn to the proof of the theorem itself. Let X be a Polish space and let A\ and
A2 be disjoint analytic subsets of X. Without loss of generality assume that A\ and
67
A2 are non-empty. By Corollary 3.2.5 we can find continuous functions f,g: A f X
such that /(AO = Ai and g(N) = A2. Arguing by way of contradiction, suppose
that A\ and A2 cannot be separated by Borel sets. For each positive integer k and
ni, n2, •••, fik G N, let the set
N{ni, n2, nk) — {111 £ Af\mi = ni for i = 1,2,..., fc}.
Clearly A\ = Umi=i /(-^(mi)) anc^ = Um=i 9(N(n>i))-
By assertion (b) above, we can choose n i ,mi G N such that f(N(m\)) and
g(N(n1)) cannot be separated by Borel sets. Likewise since
OO
f(N(mi))= ( J /(iV(mi,m2)) 1712 = 1
and OO
g(N(ni)) = [ J g(N(n1,n2))
we can choose M2,^2 G N such that /(AT{TYI\, ^2)) g(iV(rii, ^2)) cannot be
separated by Borel sets. Similarly we can choose sequences m = {mt)t and n = (nt)t
such that for each k the sets /(A r(mi,m2,...,m) !)) and g(N(ni, n2,..., nk)) cannot
be separated by Borel sets. Suppose that / ( m ) ^ <7(11). Choose open balls U and
V such that / ( m ) G U, g(n) G V, and UC\V = 0. Since / is continuous there exists
a positive integer k] such that
i\r(mi,ra2,...,mA;1) C /_1(77)
and since g is continuous there exists a positive integer k2 such that
N(n\, n2,..., nfc2)
Without loss of generality suppose that k\ > k2. Then
N(rii,n2,...,njfcj) C iV(ni,n2, - ..,nk2) Q g 1(U).
68
This imples that f(N(m i , . . . ,m f c l ) ) and #(iV(ni, ...,nkl)) can be separated by the
Borel sets U and V. This is a contradiction. Therefore it must be that / ( m ) = fif(n).
However since / ( m ) £ f{N(mi)) C Ai and g(n) € fiWm)) C A2 we have that
Ai n A2 ^ 0. Again we have a contradiction and therefore we must conclude that
A\ and A2 can be separated by Borel sets. •
Corollary 4.1.2. Let X be a Polish space, and let be disjoint analytic
subsets of X. Then there are disjoint Borel subsets of X such that An C
Bn holds for each n.
Proof Let n G N. By Theorem 3.1.2, {jm^nAm is analytic. Since An and
Um^ nAm a f e disjoint analytic sets by Theorem 4.1.1 we can choose a Borel set
Cn such that An C Cn and [jm^nAm C Ccn. Now define the Borel sets BUB2,...
by letting Bn = C„ — Cm). Then each Bn is Borel and the sets Bi, B2,
are pairwise disjoint. If a; 6 An it is clear that x € Cn. Suppose m ^ n, then
Cm C O i^mAj. Since x $ Acn and n / m we have that x Cm- Therefore x G Bn
and An C Bn as desired. •
Corollary 4.1.3. Let X be a Polish space, and let A be a subset of X. If both A
and Ac are analytic} then A is Borel.
Proof. According to Theorem 4.1.1 there are disjoint Borel subsets B\ and B2 of X
such that A C Bi and Ac C B2. It follows immediately that A — B\ and Ac = B2*
Consequently A is Borel. •
The above corollary, along with Corollary 3.4.7 which gives us a non-Borel ana-
lytic set, implies that the complement of an analytic set is not necessarily analytic.
Proposi t ion 4,1.4. Let X and Y be Polish spaces, let A be a Borel subset of X}
and let f be a function from A to Y. Then f is Borel measurable if and only if its
graph is a Borel subset of X x Y.
69
Proof. Suppose that / is Borel measurable. By Proposition 2.5.2, the graph of / is
a Borel subset of A x Y. By Lemma 3.2.2 there is a set B x C in B(X x 7 ) such
that
gr( / ) = (B x C) fl (A x F) .
Since by Proposition 2.5.1 A x Y is a Borel subset of X x Y, we have gr( / ) is a
Borel subset of X xY.
Now suppose that gr( /) is a Borel subset of X x Y , and that B is a Borel subset
of Y. Then gr(/) D {X x B) and gr( / ) 0 (X x Bc) are Borel and hence analytic
subsets of X x Y. Thus the projections of these sets on X are analytic. But these
projections are / _ 1 ( 5 ) and f~l(Bc) respectively. Furthermore the sets f~l{B) and
f~l(Bc) are disjoint, and so by Theorem 4.1.1 there are Borel sets and B2 such
that / _ 1 ( B ) C Bu f-^B0) C B2, and Bx n B2 = 0. Let x G Bx fl A. Then x G A
and f(x) G B. Consequently x G f l { B ) . This gives us that f~l(B) = Ar\Bx.
Since A and Bx are Borel, f~l{B) is Borel. Therefore / is Borel measurable. •
Proposi t ion 4.1.5. Let X and Y be Polish spaces, let A be a Borel subset of X,
let f : A Y be Borel measurable, and let B = f{A). If f is injective and if
B G B(Y), then f - 1 is Borel measurable.
Proof. Define h such that h(x, y) = (y, x). Clearly h is a homeomorphism from 1 x 7
onto Y x X. Let (y,x) G h(gv(f)). Then x G X and f(x) = y. Since / is injective
we have that f~1(y) — x. Consequently (y, x) G gr ( / 1 ) and h(gv(f)) C gr ( / ). A
similar argument shows the converse and consequently g r ( / 1 ) = h(gr(f)). Hence
g r ( / _ 1 ) is a Borel subset of Y x X if and only if gr( / ) is a Borel subset of 1 x 7 .
By Proposition 4.1.4 we have that f"1 is Borel measurable. •
We now show that the assumption that f(A) belongs to B(Y) can be removed
from Proposition 4.1.5. First we need the following lemma.
70
L e m m a 4.1 .6 . Let X and Y be Polish spaces, let A be a non-empty Borel subset
of X, and let f : A Y be Borel measurable and injective. Then there is a Borel
measurable function g \ Y —¥ X such that g{Y) C A and such that g{f{x)) %
holds at each x in A.
Proof. Let d be a metric for X , and let x be a fixed element of A. For each positive
integer n we define a function gn : Y —> X as follows. Choose a finite or countably
infinite partition {An,k}k of A into non-empty Borel subsets of diameter at most \
and in each An,k choose a point xri)k. We can construct this partition as follows.
Let n G N and let (xk)k be a sequence whose terms form a dense subset of A. Let
Ana = Bd{xu i ) n A. Let An,2 = (Bd(x2, i ) n A) - An,!. Similarly let
j J
An,j = ^ — U -^n,k-
Without loss of generality we can assume that each of these sets is non-empty,
because we could discard any set which was empty and renumber. The sets f(An,k),
k = 1,2,.. . , are disjoint and analytic by Proposition 3.2.3. Therefore by Corollary
4.1.2 we can choose disjoint Borel sets Bn,k, k = 1,2,... , such that f(An,k) C Bn,k
holds for each k. Now let gn{y) = xn,k if y G Bn,k and let gn(y) = x if y £
(Ufc Bn,k)c- Let B be a Borel set in X. Then ^ ( - B ) is the union of those Bn,k
such that xn,k G B and possibly (lj fe Bn,k)c if x G B. Therefore g'^iB) is Borel
and gn is Borel measurable. Define g : Y -4 A by letting g(y) = limn gn(y) if the
limit exists and belongs to A, and letting g(y) = x otherwise. Proposition 2.5.8 and
the statement which follows Proposition 2.5.8 imply that g is Borel measurable. If
x G A, then d(x,g„(f(x))) < ^ holds for each n. Consequently g(f(x)) = x. Thus
g is the required function. •
T h e o r e m 4.1 .7 . Let X and Y be Polish spaces, let A be a Borel subset of X, and
let f : A —>• Y be Borel measurable and injective. Then f(A) is a Borel subset ofY.
71
Proof. Assume without loss of generality that A ± 0 . According to Lemma 4.1.6
there is a Borel measurable function g :Y X such that g(Y) C A and such that
g(f(x)) = x holds at each x in A. We will now show that
/(A) = {ye Y\f(g(y)) = y}.
Let x e /(A). Then /_ 1(a;) G A and consequently
5(/(/_1 (®)) = 9(x) = /_ 1 (®).
This gives us that f{g(x)) = x and z G {y € Y\ f(g(y)) = y}- Now suppose
f(g(y)) = y. Then g(y) = f~l{y), giving f~~l{y) € A and y € /(A). Therefore
/(A) = {y € Y\ f(g(y)) = y}.
Thus Proposition 2.5.9 applied to the functions y y and f(g{y)) implies that
f(A) is a Borel subset of Y. •
4.2 Borel Isomorphisms
Let (X,A) and (Y,B) be measurable spaces. A bijection / : X Y is an
isomorphism if f is measurable with respect to *4 and B and / 1 is measurable
with respect to B and A* Equivalently, the bijection / is an isomorphism if the
subsets A of X that belong to A are exactly those for which f(A) belongs to B.
The spaces {X^A) and (Y,B) are isomorphic if there exists such an isomorphism.
We shall also call subsets X0 and Y0 of X and Y isomorphic if the spaces (X0, Ax0)
and (Y 0 lBy o) are isomorphic. In case (X,B(X)) and (Y,B(Y)) are Polish spaces
together with their Borel cr-algebras, we shall often use the term Borel isomorphism
instead of isomorphism.
Theo rem 4.2.1. Let A and B be Borel subsets of Polish spaces. Then A and B
are Borel isomorphic if and only if they have the same cardinality. Furthermore}
72
the cardinality of each uncountable Borel subset of a Polish space is that of the
continuum.
Proof. It is clear that if A and B are isomorphic then they have the same cardinality.
Suppose that A and B have the same cardinality. If these sets are finite or countably
infinite, then each of their subsets is a Borel set and each bijection between them
is an isomorphism; hence A and B are isomorphic.
Now suppose that A and B are uncountable. Then Proposition 3.4.2 implies that
there are continuous injective maps / : Af A and g : Af —»• B such that A — f(Af)
and B - g(Af) are Borel sets. Consequently f(Af) and g(Af) are also Borel sets.
Since / and g are continuous, / and g are Borel measurable. By Proposition 4.1.5, /
and g are Borel isomorphisms of Af onto f(Af) and g(.Af) respectively. Thus go f
is a Borel isomorphism of f(Af) onto g{Af). Now let I be a countably infinite subset
of / (Af) and let h be a bijection of the countably infinite set IU ( A - f (Af)) onto the
countably infinite set ( 5 ( / _ 1 ( ^ ) ) ) U (B - g{Af)). Let k be the function that agrees
with g o f-1 on f(Af) - I and with h on / U {A - f{Af)). Clearly k is a bijection
from A onto B. Let H be a Borel subset of A. Then
k(H) = k{H n (/(AO - I)) U k(H n (I U (A - /(A/"))))
= g o r 1 (H n (/(AO - 1 ) ) U h(H n (I u (A — /(Af))).
We have that f{Af), H, and I are Borel; therefore since g o f 1 is a Borel ismorphism
g o f - ^ H n i f i A T j - I ) )
is also Borel. Since h(Hn(IU(A-f{Af))) is countable, it is also Borel. Consequently
k(H) is Borel.
Now suppose J is a Borel subset of B. Then
k - \ j ) = k~\j n k{f(Af) -1)) u k-\J n k{i u (A- f(Af)))
73
= (9 o r ' T V n k{f{M) - I)) U h-\J n (g{f ^J)) u {B - g{N))).
Since gof-1 is a Borel isomorphism, from above we have that k maps Borel sets to
Borel sets, and the sets f ( A f ) — I a n ( l J are Borel, (g o / ) (J H k[f(M) I))
Borel. Since h~\Jf\{g{f~l (I)P(B - g{Af))) is countable it is also Borel. Therefore
krl(,J) is Borel. Therefore k is a Borel isomorphism from A onto B. In particular
each uncountable Borel subset of a Polish space is Borel isomorphic to R, and so
has the cardinality of the continuum. •
It follows from Theorem 4.2.1 that a Borel subset of a Polish space is Borel
isomorphic to R, to the set N of all positive integers, to the set {1,2,..., n} for some
positive integer n, or to the empty set. In Corollary 3.4.7 we showed that there is
an analytic subset of Af which is not Borel. The following proposition gives us that
each uncountable Polish space has an analytic subset which is not Borel.
Proposition 4.2.2. If X is an uncountable Polish space, then there is an analytic
subset of X that is not Borel.
Proof. By Theorem 4.2.1 each of M and X has the cardinality of the continuum and
therefore Af and X are Borel isomorphic. By Corollary 3.4.7 there is an analytic
subset S of J\f that is not a Borel subset. Let / be an isomorphism from M onto X.
Then f(S) is not a Borel subset of X. There exists a continuous function g and a
Polish space Z such that g : Z -»> M and g{Z) = S. Then ( / o g)(Z) = f(S) giving
that f(S) is analytic. Consequently there is an analytic subset of X that is not a
Borel set. •
CHAPTER V
THE MEASURABILITY OF ANALYTIC SETS
Let (X, A) be a measurable space, and let ^ be a measure on A. The measure ft
(or the measure space (X,A,fJ.)) is complete if the relations A £ A, (J-{A) = 0, and
B C A together imply that B G A. The completion of A under is the collection
A^ of subsets A of X for which there are sets E and F in A such that
1) E C A C F and
2) (jl(F-E) = 0.
A set that belongs to Ap is said to be /i-measurable.
Suppose that A C I and there are sets E and F in A that satisfy (1) and (2).
Then fx(E) = f/,(F). Furthermore if B C A and B G A then fJ-(B) < fi(F) = fJ>(E).
Hence
fi(E) = sup{/u(j5)| B G A and B C A}.
Therefore the common value of fi(E) and fJ-(F) depends only on the set A and the
measure jj, and not on the choice of sets E and F satisfying (1) and (2). Thus we
can define a function Ji : Afl —>• [0, +oo] by letting ji{A) be the common value of
jj,(E) and /J,(F) where E and F belong to A and satisfy (1) and (2). This function
/J is called the completion of fx.
We also define the outer measure fx*(A) and the inner measure of an
arbitrary subset A of X by
3) /u*(A) = mi{/j,(B)\ AC B and B G A},
and
4) n*{A) = sup{ju(5)| B C A and B G *4}.
75
on Propos i t ion 5.0.1. Let (X,A) be a measurable space, and let // be a measure
A. Then A^ is a a-algebra on X that includes A and ~p xs a measure on A^ that
is complete and whose restriction to A is fi. Furthermore /J is the only measure on
A^ that agrees with jj, on A.
Proof. It is clear that A? includes A and hence that X € A ft- Note that the relations
E C A C F and jx(F - E) = 0 imply that Fc C Ac C Ec and fj,(Ec - Fc) = 0;
thus Ap is closed under complementation. Now suppose that (An)n is a sequence
of sets in AFor each n choose sets En and Fn in A such that En C An C Fn and
fi(Fn - En) = 0. Then U n E n and U n F n belong to A and satisfy
UnEn C UnAn C Urai^n-
Also,
/ i (U nF n - U n E n ) = fi{\JnFn n ( U n E n ) c )
= /u ( (u n F n ) n ( n „ ^ ) )
= /u(un{Fn n (n„££)))
< /u(Un{Fn n Ecn))
— )J,(Un(Fn - En))
< - En) n
= o.
Thus U n A n belongs to A? and Ap is a cr-algebra on X that includes A.
Clearly Ji is an extension of /', and ]J takes on non-negative values with ^(0) = 0.
Let (An)n be a disjoint sequence of sets in A^ and for each n again choose sets En
and Fn in A that satisfy En C An C Fn and fi(Fn — En) = 0. Then the disjointness
of the sets An implies the disjointness of the sets En. Therefore
^(Unj4.ji) = /i(Un-E'n) ^ y jJ-^En) ^ ^ fiAn.
76
Thus /I is a measure. Now suppose B is a subset of X such that there exists A £ A
with B C A and fi(A) = 0. Then 0 C B C A with fi(A - 0) = 0. Therefore B is
~p-measurable and Ji is complete.
Suppose v is a measure on A^ that agrees with fj, on A. Let B G A^. If B € A
then
-p(B) = v{B) =
Now suppose that B A. Choose sets E and F in A such that E C B C F and
— E) = 0. Then u(F — E) — 0 which implies that v{F) — v{E). Since v must
be monotone it must be true that
u(E) = n(E) = u{B).
Consequently,
u(B) = n(E) = -p(B).
Now we have that JI is a unique extention of /j, to A d
Propos i t ion 5.0.2. Let (X,A) be a measurable space, let fi be a measure on A
and let A be a subset of X such that /u*(A) < oo. Then A belongs to A^ if and only
if H*(A) — fJ>*{A).
Proof. If A belongs to AM then there are sets E and F that belong to A and satisfy
E C A C F and (i{F - E) = 0. Then
Since fjt(E) = /j,(F), the relation yU*(A) = ft* (A) follows.
Now suppose A is a subset of X with /U*(A) < oo and fJ,*(A) = ji*{A). for each
positive integer n choose a set En and a set Fn from A such that En C 4 C Fn and
77
f**(A) > fi(Fn) - J and /i*(A) < fx(En) + J . Let E = U n £ „ and F = O n F n . Then
E and F belong to A, and E C. A C F. Then
2 H(F -E)< n(Fn - En) < - .
Since n was arbitrary, ji(F — E) = 0. Therefore A E A •
L e m m a 5.0.3. Let (X, A) be a measurable space, let n be a finite measure on
(X,A) and let fi* be defined by equation (3). If (An)n is an increasing sequence of
subsets of X, then
fj,*(UnAn) = lim ft* (An). n
Proof. The monotonicity of pi* implies that lim„ fi*(An) exists and that
l i m n * ( A n ) < fx*(UnAn). n
We now only need to show that
Y\mjj*(An) > jj,*((JnAn). n
Let e > 0, and for each positive integer n, use (3) to choose a set Bn that belongs
to A such that An C Bn and
fJ'(Bn) < ( A n ) + e.
By replacing Bn with (~)<j°.nBj we can assume that the sequence {Bn\ is increasing.
By Lemma 2.5.10 we have that /j,(UnBn) = l i m n f j , ( B n ) . Therefore we have
/i*(U„An) < jj,(UnBn) = l i m n ( B n ) < limfj,*(An) + e. n n
Since e is arbitrary the proof is complete. •
78
T h e o r e m 5.0.4. Let X be a Polish space, and let fj, be a finite Borel measure on
X. Then every analytic subset of X is fx-measurable.
Proof. Let A be an analytic subset of X. We shall show that A is ^-measurable by
showing that /i*(A) = ji*(A), and we shall do this by producing, for an arbitrary
e > 0, a compact subset K of A such that /x(A') > /i*(A) — e.
We can assume that A is non-empty and let e > 0. Choose a continuous function
/ : Af —>• X such that f(Af) = A. For positive integers k and n i , n 2 , n * let
£ ( n i , n-2, ..njt) be the set of those elements m of Af that satisfy m; < n{ for i =
1 , 2 , k . We shall construct an element p = (pk)k of Af such that
/**(/(£(Pi>P2,-,Pfc))) > f**(A) ~ e
holds for each k. Let us begin by choosing the first term p\ of the sequence p. Note
that { £ ( n ) } ^ | is an increasing sequence of sets whose union is Af. Consequently
{ / ( £ ( n ) ) } ^ = 1 is an increasing sequence of sets whose union is A. By Lemma 5.0.3
/ i*(A) = lim n*(f(C(n))) n
and therefore we can pick a positive integer p\ such that
^*(/(£0i))) > - e.
Since C(pi) = U r e(£(pi ,n)) , a similar argument provides a positive integer p2 such
that
V*{f(£{Pi,P2))) > P*(A) - e.
Continuing in this way we obtain a sequence p = {pk)k of positive integers such
that
^*(/(£(Pl,P2, -,Pfc))) > V*(A) - €
79
holds for each k. Now let L = Difc£(pi,p2? a n ( i ^ ^ — / ( ^ ) - Then
L = {m £ < p*, for each z} = € N|rrii < pi}. i
Since each set {m; G N|m; < p;} is finite, it is also compact. Tychonoff's Theorem
implies that L is also compact. Since a continuous image of a compact space is
compact, we have that K is a compact subset of A. We need to show that p ( i i ) >
p*(A) — e.
Let us begin by showing that K = fl £./(£&), where for each k we have abbreviated
C(pi,p2,...,pk) by Ck- It is clear that K C flfc/(£fc). Let d be a metric for the
topology on X. If x G n&/(£fc) then for each k we can choose an mfc = of
Ck such that d(f(mk), x) < | . Note that for each k the ith components of the terms
of {mfc}fc form a bounded subset of N; hence the terms of {m^} form a relatively
compact subset of Af. Consequently we can choose a convergent subsequence of
{mfc}. Let 1 be the limit of this subsequence. Then /(1) = x because we can
choose a k such that the distance from / ( m ^ ) and / ( l ) are arbitrarily small because
of the continuity of / and also we can choose k such that d(f(m.k ),ar) < \ is
arbitrarily small. Now fix k. Suppose 1 ^ Ck, then for some i < k, li > pi. But
for each j, rriji < pt. This contradicts the fact that some subsequence of {m?); }JL,
converges to Consequently m G Ck and m G C\k£>k- Hence x G K. Consequently
K = Hjkf{Ck) as desired. For each k the set f(Ck) is closed and includes /(£&).
Therefore
H(K) - n(r\kf{Ck)) - l i m n f { C k ) > H*(A) - e. k
Since fj,*A > fi*(K) > /i * (A) — e and e is arbitrary Clearly
and therefore we have equality. Consequently A is ^-measurable. •
Let (X, *4) be a measurable space. Recall from the introduction that a subset of
X is universally measurable with respect to (X, A) if it is //-measurable for every
80
finite measure (j, on (X,A). Let A* be the family of all universally measurable
subsets of X. Then A* = Hp Ap where fi ranges over the family of finite measures
on (X, A); hence A* is a cr-algebra. Also for every finite measure /J, on (X, .4) there
is a unique measure on (X,A*) that agrees with /x on A.
Now suppose that X is a Polish space. The universally measurable subsets of X
are those that are universally measurable with respect to (X,B(X)). Theorem 5.0.4
can be reformulated as follows.
Corollary 5.0.5. Every analytic subset of a Polish space is universally measurable.
Proof. The corollary is simply a restatement of Theorem 5.0.4. •
Using the above corollary and Proposition 4.2.2 we can now conclude that each
uncountable Polish space contains a non-Borel subset which is universally measur-
able.
REFERENCES
Cohn, Donald L. Measure Theory. Stuttgart : Birkhauser, 1980.
Kuratowski, K. Topology, vol. 1. New York: Academic Press, 1966.
Munkres, James. Topology: A First Course. Englewood Cliffs: Prentice-Hall, 1975.
Royden, H. L. Real Analysis. 3rd ed. New York: Macmillan, 1963.
Top Related